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# 468 Class Note for MATH M0070 at Purdue

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Date Created: 02/06/15

Course Notes for MA 460 Version 3 Jim McClure 1 De nitions and Basic Facts The goal of this course is to help you become expert in geometry so that you can teach it with con dence and pleasure We begin more or less where you left off in high school and then advance rather quickly to a higher level Here s the big picture The goal of geometry is to know the truth about geometrical gures just as the goal of astronomy is to know the truth about planets and stars and the goal of chemistry is to know the truth about atoms molecules and chemical reactions To achieve this goal we must have a way of knowing which statements are true and which statements are false In sciences like astronomy and chemistry the way to do this is to test each statement against careful observations But this method is not foolproof because no observation can be perfectly accurate For example Newton7s laws in physics were tested by thousands of observations over almost 200 years but the small uncertainties in these observations concealed the fact that there is a more accurate and fundamental theory than Newton s namely Einstein s theory of relativity Mathematics is the only science in which there is a way to achieve complete certainty with no possibility of errors due to observation The way mathematicians do this is to begin with a few very simple statements called Basic Facts or Axioms that can be accepted as true on the basis of intuition These are then used to prove everything else This means that as long as the Basic Facts are correct everything else is guaranteed to be completely correct Before we can state the Basic Facts and begin to prove things it7s helpful to have some special words such as congruent or parallel that refer to geometric gures Thus we begin with a section of De nitions 11 De nitions In order to achieve complete certainty in our proofs we must avoid any uncertainty or ambiguity in the way we use geometric terms We need to be sure that we all mean the same thing by these terms and that we re always using them in the same way The way to accomplish this is to begin with precise de nitions I assume that you already know the meanings of some simple terms like point7 line7 line segment7 and ray Angle An angle is the gure formed by two rays which begin at the same point this point is called the vertex of the angle Straight angle A straight angle is an angle in which the two rays point in opposite directions on the same line Here is a picture You know that we measure angles in degrees But what is a degree L Degree A degree is the 180 th part of a straight angle Note we will not consider angles bigger than 180 Right Angle A 90 angle is called a right angle Perpendicular Two lines are called perpendicular if they form a right angle Triangle A triangle consists ofthree points called vertices and the three line segments which connect them called sides The vertices are not allowed to be collinear that is7 they are not allowed to all lie on the same line Congruent Triangles Two triangles AABC and ADEF are congruent written AABC g ADEF if all three corresponding angles and all three corresponding sides are equal Here is a picture C F Note when we say that AABC is congruent to ADEF we mean that the vertices match up in that order that is A matches with D B matches with E and 0 matches with Similar Triangles Two triangles AABC and ADEF are similar written AABC w ADEF if all three corresponding angles are equal Note a common mistake is to say that the de nition of similarity tells you that the corresponding sides are proportional This is not part of the de nition it is Basic Fact BF 4 in Section 12 Here is a picture Parallel Lines Two lines are parallel if they do not intersect Corresponding angles If two lines m and n are crossed by a third line the third line is called a transversal then the following pairs of angles are called corresponding angles7 1 and 5 2 and 6 3 and 7 4 and 8 Midpoint of a line segment The midpoint of a segment AB is the point M on the segment for which MA MB Angle bisector The bisector of an angle is the line that goes through the vertex of the angle and splits the angle into two equal parts Quadrilateral A quadrilateral consists of four points A7 B7 C7 and D called vertices and the line segments AB7 BC7 CD and AD called sides Parallelogram A quadrilateral is a parallelogram if the opposite sides are parallel Note a common mistake is to say that the de nition of parallelogram tells you that the opposite sides are equal this is not part of the de nition7 it is Theorem 10 Rectangle A quadrilateral is a rectangle if it has four right angles Square A quadrilateral is a square if it has four equal sides and four right angles Comment on the de nitions of similar triangles and parallelogram As l7ve mentioned7 the de nition of similarity doesn t include the fact that corresponding sides of similar triangles are proportional7 and the de nition of parallelogram doesn7t include the statement that opposite sides are equal This means that when you want to say that similar triangles have equal angles7 you use the de nition of similarity7 but when you want to say that similar triangles have proportional sides7 you have to use Basic Fact 4 which is stated in Section 12 And when you want to say that opposite sides of a parallelogram are parallel7 you use the de nition of parallelogram7 but when you want to say that opposite sides of a parallelogram are equal7 you have to use Theorem 10 Since this makes your life more complicated7 I should explain why we do it this way As l7ve said7 our method for achieving complete certainty in geometry is to begin with Basic Facts that we can accept on the basis of intuition and then to prove everything else using only those But for this method to work7 we have to be careful to know exactly what were assuming In particular7 we must be careful to distinguish between Basic Facts which are statements about reality that we re assuming and De nitions which are agreements about the meanings of words The fact that when two triangles have equal angles they also have proportional sides is a rather surprising statement about reality7 it isn t a statement about the meanings of the words If we included this fact in the de nition of the word similar7 we would be making a hidden assumption about reality Here s an analogy that can make this clearer If we want to prove the statement that there is water on Mars7 we can t do it by including this statement in the de nition of the word Mars or of the word water We can only prove it by making observations7 for example by sending a spacecraft to Mars and nding water there This way of using de nitions is also what we do in ordinary life For example the dictionary de nition of presiden is the highest executive of cer in a modern republic 7 the de nition does not include the statement that Washington was the rst president of the United States although this is a true fact 12 Basic Facts Before we can begin to prove things we have to have as a starting point a list of intuitively clear facts that we accept without proof Here is a list of basic facts that you know from high school which will be the starting points for our proofs When we get to Euclid we will see that most of these basic facts can themselves be proved by starting from even simpler facts Euclid gives proofs of BF 1 BF 2 BF 3 BF 4 and the second part of BF 5 An important point In geometry its more satisfying to prove things than to base them on intuition Because of this we want to keep the list of Basic Facts as short as possible We shouldn7t add a statement to the list of Basic Facts if there s a way to prove it from the other Basic Facts BF 1 SSS if two triangles have three pairs of corresponding sides equal then the tri angles are congruent BF 2 SAS if two triangles have two pairs of corresponding sides and the included angles equal then the triangles are congruent Note it is possible for two triangles to have two pairs of corresponding sides and a pair of ntmincluded corresponding angles equal and still not be congruentican you draw an example of this BF 3 ASA if two triangles have two pairs of corresponding angles and the included side equal then the triangles are congruent Note as you know from high school the AAS criterion for congruence is also valid but we don t include it in the list of Basic Facts because we can prove itisee the end of Section 22 BF 4 If two triangles are similar then their corresponding sides are proportional that is if AABC is similar to ADEF then AB 7 AC 7 BC DE DF EF Note BF 4 is a one way street liit does not say that if the corresponding sides are proportional then the triangles are similar Eventually we will prove that ifthe corresponding sides are proportional then the triangles are similar BF 5 If two parallel lines I and m are crossed by a transversal then all corresponding angles are equal If two lines I and m are crossed by a transversal and at least one pair of corresponding angles are equal then the lines are parallel Note BF 5 is a two way street you can use it in either direction Here are some simpler Basic Facts which are included for completeness we will need these in order to give reasons for every step in our proofs Euclid gives proofs of BF 10 BF 11 BF 12 BF 13 BF 14 and part of BF 8 BF 6 The whole is the sum of its parts this applies to lengths angles areas and arcs BF 7 Through two given points there is one and only one line This means two things First it is possible to draw a line through two points Second if two lines have two or more points in common they must really be the same line BF 8 On a ray there is exactly one point at a given distance from the endpoint This means two things First it is possible to nd a point on the ray at a given distance from the endpoint Second if two points on the ray have the same distance from the endpoint they must really be the same point BF 9 It is possible to extend a line segment to an in nite line BF 10 It is possible to nd the midpoint of a line segment BF 11 It is possible to draw the bisector of an angle BF 12 Given a line I and a point P which may be either on I or not on I it is possible to draw a line through P which is perpendicular to Z BF 13 Given a line I and a point P not on I it is possible to draw a line through P which is parallel to Z BF 14 If two lines are each parallel to a third line then they are parallel to each other BF 15 The area of a rectangle is the base times the height 2 Some familiar theorems from highschool geome try Now we can start proving things The facts we state from now on will be called Theorems not Basic Facts Both Basic Facts and Theorems are true statements about realityithe difference between them is that a Basic Fact is something we accept without proof based on intuition while a Theorem is something we can prove by using the Basic Facts Basic Facts are also called Axioms A very important point In proving theorems we are allowed to use only three ingredients de nitions Basic Facts and theorems that have already been proved We have to be careful not to use Theorems that have not yet been proved because we have to avoid circular arguments we aren7t allowed to use one theorem to prove a second and then use the second to prove the rst 21 Angles formed by intersecting and parallel lines Before continuing we need a few more de nitions Adjacent and Vertical Angles When two lines cross four angles are formed The pairs of angles that share a side are called adjacent and the pairs of angles that do not share a side are called vertical In the picture the adjacent pairs are 1 and 2 1 and 4 2 and 3 3 and 4 The vertical pairs are 1 and 3 2 and 4 Interior and exterior angles If two lines m and n are crossed by a transversal then angles 1 2 7 and 8 are called exterior angles and angles 3 4 5 and 6 are called interior angles Alternate interior angles In the picture above7 the following pairs of angles are called alternate interior pairs 7 3 and 6 4 and 5 Theorem 1 When two lines cross a adjacent angles add up to 180 and b vatlcal angles are equal Proof Every possible picture which illustrates the theorem will look like Figure 17 so it is enough if we show in Figure 1 that 1 and 2 add up to 180 and that 1 and 3 are equal Now 11 2 ABC by BF 67 and M ABC 180 by the de nition of degree Combining and gives 6 11 2 180 which is what we were to prove for part a An exactly similar argument shows that MM 2 3 180 and we conclude that 1 3 by and algebra This is what we were to prove for part QED Figure 1 Note we always end a proof by writing QED This is an abbreviation for the Latin phrase Quod erat dernonstrandurn77 which means this is what was to be proved 7 Comment on the use of pictures in proofs We are allowed to use a picture in a proof as long as we use a typical picture7 that is7 one which has only the features guaranteed by the hypothesis7 and no special features An argument which is valid for such a picture will automatically be valid for every other picture which satis es the hypothesis Examples of the sort of thing you should be careful about if the theorem is to be valid for every rectangle7 your proof shouldn t be based on a picture of a square if the theorem is to be valid for every triangle7 your proof shouldnt be based on a picture of an isosceles triangle Our next theorern collects some useful variations on BF 5 Theorem 2 Suppose that Z and m are two lines crossed by a transversal a fl and m are parallel then both pairs of alternate interior angles are equal If at least one pair of alternate interior angles are equal then I and m are parallel b fl and m are parallel then each pair of interior angles on the same side of the transversal adds up to 180 If at least one pair of interior angles on the same side of the transversal adds up to 180 then I and m are parallel c fl and m are parallel then each pair of castcrior angles on the same side of the transversal adds up to 180 If at least one pair of castcrior angles on the same side of the transversal adds up to 180 then I and m are parallel Proof Every possible picture that illustrates the Theorem will look like Figure 2 Figure 2 Part a contains two statements and we have to prove both For the rst statement we are given that Z and m are parallel and we have to prove that the two pairs of alternate interior angles are equal We know that Q 11 15 by BF 5 We also know that M 11 14 by Theorem 1b Combining and tells us that 44 5 An exactly similar argument gives the equality of the other alternate interior pair For the second statement in part a we are given that a pair of alternate interior angles are equal lt7s enough to give the proof when this pair is 4 and 5 because the proof for the other pair is exactly similar So we are given 9Q 14 15 Theorem 1b tells us that M 14 11 Combining and tells us that 1 5 Since this is a pair of corresponding angles BF 5 tells us that Z and m are parallel For the rst statement of part b we are given that the lines I and m are parallel and by BF 5 this tells us that Q 11 15 Theorem 1a tells us that M 41 13 1800 10 Combining and tells us that 3 5 180 An exactly similar argument tells us that the other two interior angles also add up to 180 For the second statement of part b7 we are given that a pair of interior angles on the same side of the transversal adds up to 180 lt7s enough to give the proof when this pair is 3 and 5 since the argument is exactly similar if it is the other pair So we are given i 3 5 180 Theorem 1a tells us M 41 3 180 Combining and we see that 1 5 and by BF 5 we conclude that Z and m are parallel The proof of part c is a homework problem QED A convenient shortcut When one part of an argument repeats an earlier part word for word7 with only the names of the points and lines changed7 you7re allowed and encouraged to skip the repetition and just say this part is similar 7 This happened in the proof of Theorem 1 and several times in the proof of Theorem 2 But the part that s skipped must be an exact repetition of an earlier partiif it merely resembles an earlier part you still have to give it in full 22 The sum of the angles of a triangle Theorem 3 The angles of a triangle add up to 180 Proof Refer to Figure 3 BF 13 allows us to draw a line m through 0 which is parallel to AB Now i 41 2 3 180 by BF 6 Theorem 2a tells us that M 1 1A and x x as 3 AB Combining x7 and as x as we see that A AB 2 180 7 which is what we were to show QED 11 Figure 3 Note We dont need to consider the different ways the picture might look because the argument works exactly the same way for all possible pictures Theorem 4 If two triangles ABC and DEF have A 1D and AB 1E then also 0 1F Proof We know from Theorem 3 that 1A AB 0 180 Combining this with the given7 we have AD E 0 180 But using Theorem 3 again we have M D E 1F 180 By 9Q and algebra we conclude that AC 1F QED Comment on the AAS criterion for congruence Suppose that you have two triangles and you know that two pairs of corresponding angles and a nonlncluded pair of corre sponding sides are equal You cant apply BF 3 directly to this situation7 but Theorem 4 implies that all corresponding pairs of angles are equal7 and then BF 3 does apply So when you are in the AAS situation7 you may conclude that the triangles are congruent7 with the justi cation Theorem 4 and BF 377 For an exarnple7 see the proof of Theorem 5 12 23 Isosceles triangles Theorem 5 a If two sides of a triangle are equal then the opposite angles are equal b If two angles of a triangle are equal then the opposite sides are equal C Comment This theorem is a two way street it says that if you are given either one of the statements two sides are equal or two angles are equal then the other statement must also be true There is a convenient abbreviation for this kind of situation if we have a theorem that says if A is true then B is true and if B is true then A is true we can state it more brie y by saying A is true if and only lfB is true or even more brie y by A ltgt B For example we can restate Theorem 5 as Two sides of a triangle are equal ltgt the opposite angles are equal lt7s important to note that when we want to prove a theorem that has ltgt we have to give two proofs one for each direction Proof of Theorem 5 We begin with part a This means that in Figure 4 we are given that AC BC and we want to prove 1A AB Find the midpoint M of AB which we are allowed to do by BF 10 and connect it to C In triangles AMC and BMC we have AC BC given AM MB de nition of midpoint and MC MC Now by BF 1 we have AAMC E ABMC and from this we conclude that A AB de nition of congruent triangles l3 Figure 4 For part b we are given that A AB and we want to prove that AC BC Draw the perpendicular line m from C to AB allowed by BF 12 and give the intersection of m and AB the name D see Figure 5 Then 1 2 since both are right angles by the de nition of perpendicular and 1A AB given Thus 3 4 by Theorem 4 Furthermore CD DC so AADC g ABDC by BF 3 and from this we conclude that AC BC de nition of congruent triangles QED Figure 5 Note In this proof M and D are actually the same point However because they are constructed by different recipes the information which is available for use is different in the two parts ofthe proof for part a we are allowed to use the fact that AM MB whereas for part b what we are allowed to use is that ADC and ABDC are right angles There is a standard name for the kind of triangle described in Theorem 5 14 Isosceles A triangle with two equal sides is called isosceles The word isosceles is a Greek word meaning equal sides Using this terminology7 Theorem 5 says that a triangle is isosceles ltgt it has two equal angles More about de nitions As l7ve mentioned7 a frequently asked question about de nitions is why they don t include more information For example7 why doesnt the de nition of isosceles say that an isosceles triangle has both two equal sides and two equal angles At rst sight7 this might seem to make Theorem 5 unnecessary But more careful thought shows otherwise Suppose we did de ne isosceles to mean two equal sides and two equal angles We would still want to know that whenever two sides are equal we are guaranteed that two angles will be equalithat is7 we would want to know that if a triangle has two equal sides then it is isosceles in the new sense And we would also want to know that if it has two equal angles then it is isosceles in the new sense In other words7 we would still want to know both parts of Theorem 57 and the proof of Theorem 5 wouldn t be any easier than before The moral of this discussion is that by making the de nition more complicated we wouldn t actually have made anything else simpler So we might as well at least make the de nition as simple as possible7 and this is what mathematicians usually do To put it another way7 Theorem 5 is a fact about reality7 so our only choice is to prove it or add it to the list of Basic Facts We aren t allowed to hide it inside a de nition 24 The area of a triangle Our goal in this section is to prove that the area of a triangle is one half of the base times the height We begin with a special case Theorem 6 In triangle ABC if B is a right angle then the area of the triangle is AB BC Proof Using BF 137 draw a line m through A parallel to BC and a line n through C parallel to AB see Figure 6 Give the intersection of m and n the name D Then 1 4 by Theorem 2a using the fact that AD and BC are parallel and 2 3 by Theorem 2a again this time using the fact that AB and CD are parallel Furthermore7 AC AC7 so by BF 3 we have AABC g ACDA Now the area of ABCD is equal to the sum of the areas of ABC and CDA by BF 67 and the areas of these two triangles are the same since they are congruent so we conclude that the area of ABCD is twice the area of ABC in other words7 area of ABC area of ABCD Next we want to show that ABCD is a rectangle We are given that B is a right angle7 and so by Theorem 2b we know that ABCD is also a right angle Using Theorem 2b 15 again7 this implies that AD is a right angle7 and this in turn implies by one more use of Theorem 2b that DAB is a right angle Now we have shown that ABCD is a rectangle de nition of rectangle and so we know by BF 15 that M area of ABCD AB BC Combining and gives us the formula which was to be proved QED Figure 6 Next we need some de nitions Distance from a point to a line The distance from a point P to a line rn is de ned to be the length of the line segment from P to rn which is perpendicular to rn p Base and height of a triangle ln triangle ABC7 any side can be chosen as the base Once we have chosen the base7 the height is the distance from the remaining vertex to the line containing the base For example7 if we choose BC as the base7 then the height is the distance from A to 3 Remember that distance 7 has just been de ned to mean the perpendicular distance The perpendicular from A to BC will be inside of the triangle if the base angles are both less than 90 but if one of them is bigger than 90 it will be outside the triangle7 as shown in the following picture 16 Theorem 7 The area of a triangle is one half of the base times the height Note For each triangle we are really getting three formulas for the area because there are three ways of choosing the base In the proof of Theorem 7 for the rst time we have to consider dz erent ways that the picture might look because the argument will be different Proof of Theorem 7 Its enough to give the proof when AB has been chosen as the base the other two possibilities are proved by exactly similar arguments Draw the line gt m that goes through 0 and is perpendicular to AB as we may by BF 12 and give the intersection of this line with A B the name D There are four cases to consider Case D is between A and B Case ii D is to the right of B or to the left of A Case iii D is the same point as A Case iv D is the same point as B Case Case ii C D 17 D Case iii BD Case iv In all four cases CD is the height of the triangle by de nition of height and the angle formed by CD and line AB is a right angle by de nition of perpendicular In case we have area of ABC area of ADC area of BDC by BF 6 1 EAD DC BD DC by Theorem 6 1 AD BDDC by algebra 1 iAB DC by BF 6 and this is what was to be shown for this case In case ii7 the area of ADC is equal to the area of ABC plus the area of BDC by BF 6 and so area of ABC area of ADC 7 area of BDC 1 1 EAD DC 7 EBD DC by Theorem 6 1 2AD 7 BDDC by algebra AB DC by BF 6 and algebra and this is what was to be shown for this case In case iii A must be a right angle7 so we can apply Theorem 6 Case iv is similar QED 25 The Pythagorean Theorem and the Hypotenuse Leg The OI39EHL Theorem 8 Pythagorean theorem In a right triangle the sum of the squares of the two legs is equal to the square of the hypotenuse 18 Proof In Figure 77 we are given that ACB is a right angle7 and we want to prove that a2 62 02 where a7 6 and c are the lengths of B07 AC and AB respectively We draw a perpendicular line I from C to AB which we are allowed to do by BF 12 and label the intersection of Z and AB by F Let d and 8 stand for the lengths of AF and BF note that dec by BF 6 Now 1 ACB because both are right angles and 1A 1A so by Theorem 4 and the de nition of similarity we see that AABC w AACF Using BF 4 we see that 06 bd and algebra tells us M 62 Cd An exactly similar argument shows that AABC w ACBF7 and BF 4 tells us that ca ae which by algebra gives 6 a2 08 Combining Os and 6 as we see that a262 cecdced c cc2 and this is what we wanted to show QED Figure 7 From the Pythagorean Theorem it is easy to prove the Hypotenuse Leg criterion for congruence 19 Theorem 9 Hypotenuse Leg Theorem In triangles ABC and DEF if A and D are right angles and ifBC EF and AB DE then AABC E ADEF That is if two right triangles have the hypotenuse and a leg matching then they are congruent C A B D E Proof The Pythagorean Theorem says that A32 A02 302 and DE2 DF2 EF2 Combining these equations with the given we see that AC2 DFQ and since AC and DF are positive numbers this implies AC DF Now BF 2 tells us that AABC g ADEF QED 26 Parallelograms Theorem 10 If ABCD is a parallelogram then opposite sides of ABCD are equal Proof In Figure 8 we are given that ABCD is a parallelogram Connect A and C by a line segment Now AB is parallel to CD by de nition of parallelogram s0 11 14 Theorem 2a Also AD is parallel to BC de nition of parallelogram s0 M 12 13 Theorem 2a Furthermore AC AC and s0 AACD g ACAB BF 3 From this we conclude that AB CD and AD BC de nition of 2 QED 20 Figure 8 Theorem 11 If ABCD is a parallelogram then opposite angles of ABCD are equal Proof By Theorem 2b7 1A 1B 180 and 1B AC 180 Combining these two equations gives 1A AC The proof that B 1D is similar QED Theorem 12 If a quadrilateral has a pair of sides which are equal and parallel then it is a parallelogram The proof of Theorem 12 is a homework problem Theorem 13 A quadrilateral is a parallelogram ltgt the diagonals biscct each other that is ltgt the intersection of the two diagonals is the midpoint of each diagonal The proof of Theorem 13 is a homework problem 3 More about triangles 31 The line through the midpoints of two sides of a triangle We begin with a convenient piece of algebra Theorem 14 a Suppose that C is a point on the segment AB C is the midpoint of AB ltgt AB 2A0 b A line segment can have only one midpoint Proof For a7 we begin with the gt direction7 so we are given that C is the midpoint of AB Now AB A0 BC by BF 6 and also AC BC de nition of midpoint Combining these equations gives AB 2A0 In the lt direction7 we are given that AB 2A0 But we also have AB ACBC by BF 6 Combining these equations gives AC BC7 s0 0 is the midpoint of AB de nition of midpoint 21 For b7 suppose that C and C were both midpoints of AB Then AC AB and AC AB both by part a and so AC AC Now BF 8 tells us that C and C are the same point QED Theorem 15 In triangle ABC let D be the midpoint of AC and suppose that E is a point on BC with DE parallel to AB Then E is the midpoint of BC and DE ABi Proof See Figure 9 By BF 57 1 2 and 3 4 Also7 AC 0 Thus AABC w ADEC de nition of similarity7 and therefore AC 7 BC 7 AB l DC EC DE by BF 4 Since we are given that D is the midpoint of A07 we can apply Theorem 14a to get AC M m2 BC Combining and we see that 7 27 so that E is the midpoint of BC by EC Theorem 14a Finally7 and also tell us that AB 72 DE so that DE AB QED Figure 9 Our next theorem is closely related to Theorem 15 but considerably trickier to prove 22 Theorem 16 In triangle ABC let D be the midpoint ofAC and let E be the midpoint of BC Then DE is parallel to AB and DE AB Notice that the picture which illustrates this theorem is the same as that for Theorem 15 but the information we are given about this picture is different Proof Use BF 13 to draw a line in through D which is parallel to AB Give the intersection of m and BC the name F Then DF is parallel to AB so we can apply Theorem 15 to conclude that F is the midpoint of BC But then F is really the same point as E by Theorem 14b Now BF 7 tells us that DE is the same line as DF so DE is parallel to AB by Finally since we now know that E satis es the hypothesis of Theorem 15 we can apply Theorem 15 to conclude DE AB QED The proof of Theorem 16 is the trickiest we have seen so far You should study this proof carefully because we will be using similar ideas in several other proofs and homework problems Note that Theorem 16 would be rather easy if we had a Basic Fact that said that triangles with two sides proportional and the included angle equal are similar But we dont include this fact among the Basic Facts because we can prove it using the Basic Facts we already have We will do so in the next section 32 The SAS and SSS criteria for similarity Theorem 17 SAS for similarity In triangles ABC and DEF if 0 1F and A0 B0 m E then AABC w ADEF The proof is rather tricky The basic idea is to make a copy of ADEF on top of AABC and then show that the copy is similar to AABC But we wont know that the copy is actually congruent to ADEF until almost the end of the proof Proof See Figure 10 On ray 7A mark a point D with CD FD this is allowed by BF 8 Draw the line in through D which is parallel to AB allowed by BF 13 and give the intersection of m and BC the name E Our rst goal is to show that AD E C w AABC Since in is parallel to AB we can use BF 5 to see that 1 2 and 3 4 Also 0 0 so AD E C w AABC Our next goal is to see that AD E C E ADEF Because of the way D was con structed we know M D C DE 23 Next we use and BF 4 to get AC BC 9 DC EC Combining and 6 as we have 0 BC DFFEC and comparing this with the given we obtain BC 7 BC EC EF which by algebra gives E C EF But we were given that AC 1F so we can apply BF 2 to get MM AD E C E ADEF Now tells us that 1 2 and tells us that 1 1D so we conclude that 2 1D Sirnilarly7 4 1E And we were given 0 1F so AABC w ADEF by de nition of similarity QED Figure 10 There is also a criterion for sirnilarity analogous to the SSS criterion for congruence its the other direction of BF 4 AB AC BC Theorem 18 SSS for 51mllar1ty In tnangles ABC andDEF z m 7 W 7 W then AABC w ADEF The proof is a homework problem 24 33 The sine of an angle In this section we introduce the sine of an angle and use it to give a new formula for the area of a triangle First we need an of cial de nition for the sine7 and before stating it we need to set H the stage Let ABC be any angle7 and choose a point D on BC By BF 12 we can gt gt draw a line m through D perpendicular to AB Give the intersection of m and AB the name E Notice that there are three cases7 depending on whether ABC is less than 90 7 equal to 90 7 or greater than 90 c Dc c D D DD B E OA 35 A E0 B A ABC lt 90 ABC90 ABC gt 90 In all three cases the sine is given by the same formula DE De nition of sine The sine of ABC is Notice that for angles less than 90 this agrees with the formula you learned in high school opposite sin hypotenuse When applying the de nition of sine to an angle you have the freedom to choose any point D you want as long as D is on one of the sides of the angle All choices are guaranteed to give the same answer7 because if you choose D and your friend chooses DE D the triangles DEB and D E B will be sirnilar7 and so E will be equal to by BF 4 See Figure 11 DB 25 D D 0 0 B E E39 A Figure 11 There is a useful relationship between the sines of angles that add up to 180 Theorem 19 sin1ABC sin180 7 ABC Proof In Figure 127 FBC 180 7 ABC by BF 6 and algebra When we apply the de nition of sine to ABC we get DE sin1ABC E and when we apply the de nition of sine to FBC we get DE sinzFBC E so sin1ABC sinzFBC QED 0407 B E F Figure 12 Theorem 20 For any triangle ABC the area can be calculated by any of the following three formulas area of AABC AB AC sin A 26 area of AABC AB BC sin 1B area of AABC AC BC sin 40 that is the area is one half the product of two sides times the sine 0f the included angle Proof It is enough to give the proof of the rst formulaithe proofs of the other two are completely sirnilar There are three cases A lt 90 7 1A 90 7 and LA gt 90 see Figure 137 but in fact the proof is the same in all three cases Let us choose AB to be the base ofthe triangle7 and as usual use BF 12 to draw a line through 0 perpendicular gt gt to AB which intersects AB at a point E Then CE is the height of the triangle7 and so we have 1 90 area of AABC AB CE by Theorem 7 But the de nition of sine says sin1A g and so M CE CA sin1A Combining and gives the formula we wanted to prove QED A i Figure 13 By comparing the three formulas in Theorem 207 we get an interesting relationship called the Law of Sines Theorem 21 Law of Sines In any triangle ABC sin1A 7 sinzB 7 sinzC B0 A0 AB 27 Proof Theorem 20 tells us that l l EAB AC sin1A EAB BCsinzB since both sides of the equation are equal to the area of AABC Now and algebra give sin1A sinzB 30 AC The proof of the other equality is completely similar QED 28 4 Concurrence Theorems De nition of concurrent lines Three lines are concurrent if they meet at a single point concurrent lines nonconcurrent lines Its very unusual for three lines to be concurrentiordinarily three lines will form a triangle But in this chapter we will see that certain kinds of lines associated with a triangle are forced to be concurrent 41 Concurrence of the perpendicular bisectors7 angle bisec tors7 and altitudes Our rst concurrence theorem77 concerns the perpendicular bisectors of the sides of a triangle De nition of perpendicular bisector The perpendicular bisector of a line segment is the line that goes through the midpoint and is perpendicular to the segment A common mistake Suppose you have a point A and a line segment BC7 and you put the following statement in a proof Draw the perpendicular bisector of BC through A77 This is wrong because the perpendicular bisector of BC may not go through A Its a useful exercise to see why a step like this is not justi ed by BF 10 and BF 12 BF 10 allows us to nd the midpoint M of BC7 and BF 12 allows us to draw a line m through M which is perpendicular to BC then m is the perpendicular bisector of BC On the other hand7 BF 12 also allows us to draw a line n through A which is perpendicular to line BC But m and n will not ordinarily be the same line7 so its not allowable to say Draw the perpendicular bisector to BC through A7 because ordinarily there is no such line 29 Theorem 22 For any triangle ABC the perpendicular bisectm s of AB AC and BC are concurrent It turns out the that obvious way of trying to prove Theorem 22 leads to a dead end Anyone s rst idea would be to draw the three perpendicular bisectors and see what happens But then there would be two cases the three perpendicular bisectors might meet in a point as in Figure 14 Figure 14 but since we don t yet know Theorem 22 we cant assume that the picture wouldn t look like Figure 15 30 Figure 15 and Figure 15 is a dead end there s nothing interesting we can say about it lnstead7 we do something trickier We draw two of the perpendicular bisectors7 then draw the line segment connecting their intersection to the third rnidpoint7 and use congruent triangles to show that this line segment is the third perpendicular bisector Here are the details Proof of Theorem 22 Find the rnidpoints M7 N7 and P of AB7 A07 and BC re spectively which we are allowed to do by BF 10 Now draw a line m through M perpendicular to AB and a line n through N perpendicular to AC allowed by BF 12 and give the intersection of m and n the name X Connect X to P There are three cases Case X is inside the triangle Case ii X is outside the triangle Case iii X is on BC 1711 give the proof of Case here and ask you to do the other two on the homework See Figure 16 Our strategy is to show that XP is the perpendicular bisector of BC First observe that AM MB de nition of rnidpoint7 MX MX7 and AAMX BMX by de nition of perpendicular So AAMX E ABMX by BF 27 and hence AX BX by the de nition of congruent triangles Sirnilarly7 AANX E ACNX7 so M AX OX 31 Combining and gives BX CX But also OP BP de nition of midpoint and PX PX so ACPX E ABPX by BF 1 and therefore 9F 11 2 de nition of congruent triangles On the other hand7 Theorem 1a says that MM 11 2 180 Combining 6 as and we see that 1 90 7 so PX is perpendicular to BC de nition of perpendicular Since we already knew that P is the midpoint of BC7 we conclude that T PX is the perpendicular bisector of BC de nition of perpendicular bisector Now X is on the perpendicular bisectors of AB and AC because of the way X was constructed and also on the perpendicular bisector of BC by That is7 all three perpendicular bisectors contain X7 and so they are concurrent de nition of concurrent QED A M B Figure 16 De nition of circumcenter The point where the three perpendicular bisectors of the sides of a triangle meet is called the circumcenter of the triangle 32 Note In the proof of Theorem 227 X turned out to be the circumcenter of AABC and we showed that XA XB XC This means that the circle with center X that goes through A also goes through B and 0 Thus X is the center of a circle that goes through the three vertices of the triangle This circle is called the circumscribed circle of AABC7 and because of this its center X is called the circumcenter We conclude this section with two more concurrence theorems Theorem 23 For any triangle ABC the bisectors of 1A 1B and 0 are concurrent The proof is a homework problem De nition of incenter The point where the three angle bisectors meet is called the incenter of the triangle The incenter turns out to be the center of a circle which is tangent to all three sides of the triangle this circle is called the inscribed circle of the triangle and this is the reason for the name incenter7 De nition of altitude An altitude of a triangle is a line that goes through a vertex of the triangle and is perpendicular to the opposite side Theorem 24 For any triangle ABC the three altitudes are concurrent The proof is a homework problem De nition of orthocenter The point where the three altitudes meet is called the or thocenter of the triangle The name orthocenter 7 is somewhat misleading since this point isn t the center of an interesting circle The name comes from the fact that orthogonal is another word for perpendicular 42 Concurrence of the medians De nition of median A median of a triangle is a line that goes through a vertex of the triangle and through the midpoint of the opposite side We will show that the medians of a triangle are always concurrent In order to prove this7 we need a preliminary fact which is interesting for its own sake Theorem 25 The point where two medians of a triangle intersect is 23 of the way from each of the two vertices t0 the opposite midpoint 33 Proof In Figure 17 We are given that AP and BN are medians and that X is their intersection We need to prove that AX AP and BX BN Connect N and P N and P are the midpoints of AC and BC by de nition of median so by Theorem 16 we see that NP is parallel to AB But then 1 4 and 2 3 Theorem 2a and so AABX w APNX Theorem 4 and the de nition of similar triangles Now BF 4 implies AX BX AB W m W But Theorem 16 also tells us that 2 so implies M AX 2XP On the other hand BF 6 gives ww AXXPAP and combining and 6 we see that AX AP A similar argument shows that BX gBN QED Figure 17 Theorem 26 For any triangle ABC the three medians are concurrent Proof Find the midpoints of AB AC and BC as we may do by BF 10 and call them M N and B respectively Then AP BN and CM are the medians of AABC de nition of median 34 Give the intersection of AP and BN the name X see Figure 18 Then 2 AX EAP by Theorem 25 Next7 give the intersection of AP and CM see Figure 19 the name Y Then M AY AP by Theorem 25 Combining and we see that AX AY But then X and Y are the same point7 by BF 8 Now X is on AP and BN because of the way it was constructed and it is also on CM because X is the same point as Y So all three medians contain X7 and hence they are concurrent de nition of concurrent QED M B Figure 18 Figure 19 De nition of centroid The point where the three medians meet is called the centroid ofthe triangle 43 The Euler line De nition of collinear Three points are said to be collinear if they all lie on the same line In this section we will prove Theorem 27 Let ABC be any triangle Let 0 be the circumcenter of ABC let G be the centroid of ABC and let H be the orthocenter 0f ABC4 Then 0 C and H are collinear 35 This theorem was rst proved by Euler in the eighteenth century by analytic geometry The proof we will give was rst discovered in the nineteenth century Proof See Figure 20 We will use an indirect approach7 so we use BF 8 to construct a gt point H on 0G7 on the opposite side of G from 07 with H G 20G Our goal is to show that H is the same point as H7 and for this we only need to show that H lies on all three altitudes of AABC Construct the midpoint M of AB possible by BF 10 and connect CM CM is a median of AABC de nition of median and so G is on CM de nition of centroid Theorem 25 tells us that CG gCM Using BF 6 and algebra7 we see that MG CM7 so 0G7 M W72 Now draw in 571 and H By Theorem 1b we have i i OGM H GC Combining as7 M7 6 as and Theorem 17 we obtain MM AOGM w AH GC Now by de nition of similar triangles we have OMG H CG7 and by Theorem 2a we see that gt T OM is parallel to CH Next observe that the perpendicular bisector of AB contains 0 by de nition of circumcenter and M by de nition of perpendicular bisector It is therefore the same line as OM by BF 7 this tells us that OM is perpendicular to AB by de nition of perpendicular bisector and so we have 1 90 by de nition of perpendicular Therefore 2 90 by BF 5 This tells us that CH is perpendicular to AB de nition of perpendicular and therefore CH is an altitude of AABC de nition of altitude That is7 H is on the altitude of AABC that goes through 0 Similarly7 H is on the other two altitudes of AABC Since H is also on all three altitudes de nition of orthocenter7 we conclude that H and H are the same point gt gt Since we constructed H to be on 0G7 we now know that H is on 0G7 so 07 G and H are collinear de nition of collinear QED 36 Figure 20 37 5 The Theorems of Menelaus and Ceva 51 The Theorem of Menelaus Our next goal is to prove Theorem 28 Theorem of Menelaus Let ABC be any triangle Let A be a point gt gt of BC other than B and 0 let B be a point ofAC other than A and C and let 0 be gt a point ofAB other than A and B If A B and C are collinear then AB BC CA A C B A C B The following picture illustrates the two cases of the theorem either the line A B C crosses the triangle7 or it stays entirely outside the triangle c The Theorem is named after Menelaus7 who was a Greek mathematician in Alexan gt gt dria in the second century AD A line which crosses all three of the lines AB7 AC and gt BC7 without going through any of the points A7 B and C7 is called a Menelaus llne Proof We have to prove both cases Case The Menelaus line A B C crosses the triangle Case ii The Menelaus line ABC is entirely outside the triangle 1711 give the proof for Case i the other case is a homework problem See Figure 21 Draw a line m through B parallel to AC allowed by BF 13 and give the intersection of m with the line A B C the name X Now 1 2 and 3 4 38 both by BF 5 applied to the parallel lines AC and m7 so ABC X w AAC B by Theorem 4 and the de nition of similarity This allows us to apply BF 4 to get C B BX C A B A Next7 we have 5 6 and 7 8 both by Theorem 2a applied to the parallel lines AC and m and so ABA X w ACA B by Theorem 4 and the de nition of similarity This allows us to apply BF 4 to get M AB iBX AC Bv Solving for BX gives 9F BX gi B A and solving for BX gives A B BX MM AC B C Combining 6 as and we have GB 743 CABAiACBC and multiplying both sides of by 3 Bi gives iABHCOA 7 A C B A C B which is what we were to prove QED 39 Figure 21 Note The statement of Menelaus s theorem looks complicated but there s an easy way to remember it Suppose we want to apply it to the triangle PQR and the Menelaus line SMA in the following picture First write out a partial version of the Menelaus equation using just the points on the Menelaus line in any order S M A 1 S M A Now ll in the rest by pairing each point on the Menelaus line with the two vertices ofthe side that contains it but make sure that each vertex appears once in the numerator and once in the denominator In our example the point S gets paired with Q and R the point M gets paired with P and Q and the point A gets paired with P and R We can choose whether to put SQ or SR in the numerator but after that all choices 40 are determined If we put SQ in the numerator then in the second factor MQ has to go in the denominator so that Q will appear once in the numerator and once in the denominator and so on SQ MP AR 7 fmg l If you go through the process with the points on the Menelaus line in a different order or if you make a different choice about what goes in the numerator in the rst factor you get a different way of writing the same equation For example if we put the points on the Menelaus line in the order ASM and put AP in the numerator ofthe rst factor we get AP SR MQ 71 AR SQ MP which is just the reciprocal of the previous equation 52 The converse of a statement The word converse is often useful in understanding logical relationships in geometry Here7s what it means when you have a statement if A then B the converse of the statement is if B then A That is in the converse the given and to prove are switched The converse of a true statement may or may not be true When both the statement and its converse are true we use the symbol Here is an example from ordinary life the statement if a person lives in West Lafayette then they are an Indiana residen is true But its converse says if a person is an Indiana resident then they live in West Lafayette and this is false 53 The converse of Menelaus s theorem preliminary discus sion Now let us consider Theorem 28 and its converse We are given a triangle ABC and gt gt gt points A B and C on the lines BC AC and AB respectively Theorem 28 says that if A B and C are collinear then the equation a A B B C C A 7 AC BA 03 is satis ed That is the Theorem says that A B and C collinear gt equation is satis ed 41 The converse of Theorem 28 is the statement whenever equation is satis ed7 A 7 B and C will be collinear 7 Rather surprisingly7 this statement is not true You can see this by doing an experiment with Geometer7s Sketchpad If we choose A and B7 there will always be two points 0 for which equation is satis ed7 but only one ofthese two points will be collinear with A and B On the other hand7 the converse of Theorem 28 is close to being true7 since for each choice of A and B there is only one bad ie7 noncollinear point C which satis es equation This suggests that we should be able to get a true statement by making a small modi cation to the converse of Theorem 28 lt The rst thing to notice is that7 of the two points 0 on the line AB which satisfy equation as one of them is always inside the line segment AB and the other one is always outside of AB This suggests that its important to keep track of whether the points are on the original nonextended sides of the triangle or not Experimenting with Geometer7s Sketchpad7 we arrive at the following corrected version of the converse of Menelaus s theorem Let ABC be any triangle Let A be a point of B other than B and gt 07 let B be a point of AC other than A and C7 and let 0 be a point gt of AB other than A and B Suppose that 1 the equation AB BC CA 1 AC B A C B is satis ed and one of the following is true 2a Two ofthe points A 7 B and C are on the original nonextended sides of the triangle and the third is not or 2b None of the three points is on the original nonextended sides of the triangle Then A 7 B and C are collinear So far we have only experimental evidence for the statement in the box and we still need to prove it But there s another issue that must be dealt with rst the statement in the box is messy7 and this makes it both hard to prove and hard to use in proofs We need to nd a way to say the same thing in a more convenient way The way to do this is to incorporate a or 7 sign in ratios like to keep track of whether 0 is inside 0 the segment AB or not We will explain how to do this in the next section 42 54 Signed ratios Let us begin by thinking about points on a number line 3 2 1 0 1 2 Every point on the line has a coordinate which is a real number that speci es the exact location of the point If A and B are two points on the line with coordinates a and b then the length of AB is given by the formula la 7 bl Now notice that if we leave out the absolute value symbol we get a number a 7 b which contains two pieces of information its size is the length of AB and its sign tells whether A is to the right or to the left of B A is to the right of B when the sign is positive and to the left of B when the sign is negative Next let us consider the kind of ratios that occur in Theorem 28 So let 0 A and B be three points on the number line and let 0 a and b be their coordinates The ratio C A lc 7 al is iven b the formula 03 g y lc i bl b which contains two pieces of information its size is OB C 7 sign is described in our next theorem When we leave out the absolute values symbols we get a number and its 0 7 Theorem 29 The number C 7 negative if C is inside AB Z is positive if C is outside of the segment AB and Proof There are four cases to consider Case i C is to the right of both A and B Case ii 0 is to the left of both A and B Case iii 0 is between A and B and A is to the left of B Case iv 0 is between A and B and A is to the right of B c 7a In Case 1 076 l i H C a i i i i positive In Case 11 76 1s pos1t1ve because the numerator and denominator are C 7 both negative In Case iii 0 c is positive because the numerator and denominator are both 0 i i i i i b is negative because the numerator is positive and c 7 the denominator is negative In Case iv c a l i b is negative because the numerator in negative and the denominator is positive QED Comparing Theorem 29 with the the box on page 42 we see that it should be possible 0 76 to simplify the statement in the box by using formulas like However in the 43 gt gt gt situation were interested in the lines AB7 AC and B0 are not number lines This is not a serious problem because we can make any line into a number line by picking an origin and a positive direction This leads us to the following de nition De nition of signed ratio Let I be any line and let C 7 A and B be three points on I Make Z into a number line by choosing an origin and a positive direction and let c 7 a and b be the coordinates of C 7 A and B We de ne H C A gt gt C B 0 7 a i i i to be and we call this a signed ratio C 7 gt C A One way to think about the signed ratio H is that it is the quotient of the vector C B gt gt gt gt CA by the vector C B It isn t usually possible to divide one vector by another7 but it is possible in our situation since the vectors are on the same line and must therefore be scalar multiples of each other The signed ratio is positive if the vectors point in the same direction and negative if they point in opposite directions However7 we will not be using the theory of vectors in this course gt Another way to think about H is given by our next theorem C B H Theorem 30 The number H is equal to 03 C B if C is inside the segment AB if C is outside of the segment AB and is equal to 7 CB 1 i A C 7 0 Proof The absolute value of H is 0 al which is the same as Theorem 29 03 lo 7 bl B says that the sign of is positive if C is outside of the segment AB and is negative gt C B if C is inside the segment AB QED A question that arises in connection with the de nition of signed ratio is whether H is well de ned 7 That is7 if you and your friend are given the same line I and the OB same points C 7 A and B7 but you choose two different ways to make I into a number 44 gt line do you get the same number for H 7 The answer is yes First notice that if you move the origin from the position you chose to the position your friend chose then the numbers 0 a and b will change but the differences 0 7 a and c 7 b will not Next if your friend chose the positive direction differently from you then his values of c 7 a and 0 7 b will be the negatives of yours but these negative signs will cancel in the quotient c 7 a c 7 h We conclude this section with an important fact about signed ratios which will be used in our future work So you and your friend will get the same value Theorem 31 Letl be any line and let 0 C A and B be points on I with A not the gt gt gt gt C A C A same point as B If H H then C is the same point as 0 0 3 C B Proof Make Z into a number line and let 0 c a and 6 stand for the coordinates of C C A and B We are given gt gt gt gt C A C A gt gt gt gt C B C B and by the de nition of signed ratio this means c 7a c 7a 96 0 76 c 7b Cross multiplying in gives 0 7 ac 7 h c 7 bc 7 a and simplifying this we obtain do 7 ac 7 06 ab do 7 he 7 ca ab Cancellation gives 7ac 7 db 760 7 ca Moving the terms with c to the left and the terms with c to the right we get M c a 7 h c a 7 6 Now we are given that A and B are different points so their coordinates a and h are different numbers and therefore a 7 6 7E 0 This allows us to cancel a 7 h in M so we have This means that the coordinates of C and C are the same number so 0 and C are the same point QED Notice that Theorem 31 would not be true if we used ordinary ratios instead of signed ratios Whenever C A and B are points on a line there will always be a point H 39 l A i i OA CHA C dz erent from C for which the ordinary ratios OB and CUB are the same When this happens one of the points 0 and C will be inside the segment AB and the other gt gt gt gt C A C A will be outside so the signed ratios H and H will be different C B C B 55 The converse of Menelaus s theorem We can now restate conditions 2a and 2b in the box on page 42 in a simple and gt gt gt gt gt gt AB BC CA convenient way condition 2a says that two of the signed ratios H i and H A C B A C B are negative and condition 2b says that none of them are negative So 2a and 2b together can be combined into the single statement that the product gt gtgt gtgt gt AB BC CA i i i A C B A C B is positive And combining this with condition 1 we see that the box on page 42 simply says that if gt gtgt gtgt gt LB C A then the points A B and C will be collinear We still need to prove this But rst let us pause to check that Theorem 28 itself remains true with signed ratios lt gt Theorem 32 Let ABC be a triangle Let A be a point of BC other than B and 0 let gt B be a point offlEgtV other than A and C and let 0 be a point ofAB other than A and B If A B and C are collinear then gt gtgt gtgt gt A B BC C A i i A C B A C B Proof Theorem 28 tells us that the equation is true if we ignore the signs so we know that the left hand side is equal to either 1 or 71 and we want it to be 1 Thus all we have left to show is that the left hand side is positive There are two cases to consider 46 Case The Menelaus line A B C crosses the triangle Case ii The Menelaus line A B C is entirely outside the triangle For Case i7 see Figure 22 In this case two of the points are on the nonextended sides of the triangle and one is on an extended side7 so two of the signed ratios are negative and one is positive Thus the product of all three is positive Figure 22 For Case ii7 see Figure 23 In this case all three points are on extended sides of the triangle7 so all three signed ratios are positive and the product is also positive QED A B Figure 23 G Our next theorem is the converse of Theorem 32 this is the corrected converse of Menelaus s theorern 47 lt gt Theorem 33 Let ABC be a triangle Let A be a point of BC other than B and 0 let gt gt B be a point ofAC other than A and C and let 0 be a point ofAB other than A and B If gt gtgt gtgt gt B C A gt gt gt gt gt gt 1 C A B then A B and C are collinear lt gt Proof See Figure 24 Draw the line A B 7 and give the intersection of this line with gt AB the name 0 Then A 7 B and C are collinear We want to show that C is the same point as 0 Since A 7 B and C are collinear7 we can apply Theorem 32 to get gt gtgt gtgt gt A C B A C B Combining this with the given7 we obtain gt gt gt gt C A C A gt gt gt gt C B C B Now Theorem 31 tells us that C is the same point as 0 7 and combining this fact with we see that A 7 B and C are collinear QED cu Figure 24 48 56 The Theorem of Ceva and its converse lt gt Theorem 34 Theorem of Ceva Let ABC be a triangle Let A be a point of BC gt other than B and 0 let B be a point of AC other than A and C and let 0 be a point gt gt gt gt ofAB other than A and B IfAA BB and 00 are concurrent then gt gtgt gtgt gt A B B C C A gt gt gt gt gt gt 7 C A B C B A A C39 B The proof is a homework problem Ceva proved this theorem in the late 1600 s The converse of Theorem 34 is also true Theorem 35 Let ABC be a triangle Let A be a point ofB other than B and 0 let B be a point offlEgtV other than A and C and let 0 be a point ofA B other than A and B If gt gt gt gt gt gt A B BC CA gt gt gt gt gt gt 0 BA CB 71 lt gtlt gt lt gt then AA BB and 00 are concurrent The proof is a homework problem As an example of how useful Theorem 35 can be7 let us use it to give a new proof of Theorem 26 concurrence of the medians 49 A M Given M N and P are midpoints To prove AP BN and CM are concurrent Proof We know PB PC by de nition of midpoint so 1 Also P is between B and C so by Theorem 30 we have H PB 7 PB 7 1 1 PC 1 AZ Similarly 71 and 71 Combining these equations gives NA MB gt gt gt gt gt gt PB NC MA 1 PC NA MB gt so we may apply Theorem 35 with A P B N and C M to conclude that AP gt gt BN and CM are concurrent QED 6 Circles De nition of circle A circle consists of all of the points which are at a given distance called the radius from a given point called the center 61 Inscribed angles and central angles De nition of inscribed angle Given three points A B and C on a circle the angle ABC is said to be inscribed in the circle 50 De nition of central angle Given a circle with center 0 a central angle is an angle with its vertex at 0 If A B and C are points on a circle with center 0 there is a relationship between the inscribed angle ABC and the central angle AOC However the relationship is somewhat complicated and depends on the positions of A B C and O The following pictures show the possible cases Figure 25 Figure 26 Figure 27 Figure 28 Experimenting with Geometer7s Sketchpad shows that for pictures like Figures 25 26 1 1 and 27 ABC izAOC but for pictures like Figure 28 ABC 180 7 izAOC It turns out that in order to know which formula applies we just have to know the location of point B Theorem 36 Let A B and C be points on a circle with center 0 1 a IfB is outside of AOC then ABC EAAOC 51 b IfB is inside of AOC then ABC 180 7 zAOC Proof For part a there are four cases i O is on ray BA ii 0 is on ray BC iii 0 is inside ABC iv 0 is outside ABC For case see Figure 29 We know that OB 007 by de nition of circle7 so 1003 1030 by Theorem 5 Let us call both of these angles cc and let y AOC Then i y 4300 180 by Theorem 1a7 and M 296 1300 180 by Theorem 3 Combining equations and gives y 29 which is what we were to prove in this case Figure 29 Case ii is precisely similar to case For Case iii7 see Figure 30 We can apply case ii to ABD to get 1 i ABD iMOD and we can apply case ii to CBD to get 1 M CBD 54001 52 Finally7 we have ABC LABD ACBD by BF 6 1 1 ELAOD EACOD by and 1AOD ACOD by algebra ono by BF6 so ABC zAOC which is what we were to prove in this case Figure 30 For case iv7 see Figure 31 We can apply case ii to ABD to get i ABD zAOD and we can apply case ii to CBD to get a CBD 4001 Then we have ABC ABD 7 CBD by BF 6 and algebra zAOD 7 4001 by and l 1AOD 7 ACOD by algebra l EAAOC by BF 6 and algebra 53 so ABC zAOC which is what we were to prove in this case c D Figure 31 It remains to prove part b of the Theorem For this7 see Figure 32 We know that 0A OB7 by de nition of circle7 so OAB OBA by Theorem 5 Let us call both of these angles cc Similarly7 OBC OCB let us call both of these angles y Next we apply Theorem 3 to triangles A03 and COB to get AOB 180 7 295 and M 1003 180 7 2y Now ZAOC ZAOB 1003 by BF 6 180 7 29 180 7 2y by and 360 7 295 y by algebra 360 7 24ABC by BF 6 So we have shown that AOC 360 7 QAABC Solving for ABC gives ABC 180 7 zAOC which is what we were to prove in this case QED 54 Figure 32 Some comments on the proof Notice that in all four parts of the proof of Theorem 36 the basic strategy is to draw line OB Also notice that once case ii of part a has been proved we are allowed to use it to prove the remaining cases without having to worry about a circular argument Finally notice the general resemblance between the proof of part a and the proof of Theorem 7 Because Theorem 36 contains two different formulas which are valid in different situ ations it can be hard to remember In fact there is a way to unify the two formulas into a single formula using the idea of arc measure which we discuss in the next section 62 Arcs and arc measurement An arc is a piece of a circle We can measure an arc in degrees here is the of cial de nition Degree of arcmeasure A degree of arc on a circle is the th part of the full circle Note Although angles can never be bigger than 180 arcs can be anything up to 360 In order to use this de nition we need to have a way of naming arcs First notice that the endpoints of an arc do not determine the arc there are always two different arcs with these endpoints for example in Figure 33 there are two different arcs with endpoints AB one of which goes through 0 and the other through D In order to specify an arc unambiguously we have to give the two endpoints and a point in the middle For example in Figure 32 the two arcs with endpoints A and B are denoted ACE and ADB One thing that may cause confusion is that occasionally we will create a point on an arc whose only purpose is to help name the arc 55 Figure 33 Next we will consider the arc cut off by an angle We begin with the case of a central angle De nition of arc cut off by a central angle The arc cut off by a central angle is the part of the circle inside the angle see Figure 34 Figure 34 The most important fact about the arc cut off by a central angle is Theorem 37 The arc cut o by a central angle is the same number of degrees as the angle Proof It suf ces to show that a 1 central angle cuts of a 1 arc In Section 11 we de ned a 1 angle to be the th part of a straight angle7 so it suf ces to check that a central straight angle cuts off a 180 arc But this is true because a central straight 56 angle divides the circle into two equal pieces7 and we have de ned the whole circle to be 360 of arc QED Next we consider the arc cut off by an inscribed angle De nition of the arc cut off by an inscribed angle The arc cut off by an inscribed angle ABC is the part ofthe circle inside the angle In Figure 35 the part of the plane inside ABC is shaded7 and the arc cut off by ABC is the shaded arc Figure 35 By analogy with Theorern 377 its natural to ask whether there is a simple relation ship between the size of an inscribed angle and the size of the arc which it cuts off Experimenting with Georneter7s Sketchpad shows that the angle is always l of the arc 2 We will prove this in the next section 63 Inscribed angles and arcs We can use the idea of the arc cut off by an angle to give a sirnpler7 uni ed version of Theorem 36 Theorem 38 Let A B and C be points on a circle and let ADC be the arc cut o by ABC Then ABC arc ADC that is the number of degrees in angle ABC is half the number of degrees in arc ADC 57 0 ice a e on y r0 e 0 e pom in is eorem 1s 0 a 0w us 0 name e N t th t th l l fth t D i th th t ll t th arc ADC Proof There are four possible cases shown in Figures 36 37 38 and 39 In Figures 36 37 and 38 the arc cut off by AOC is ADC so in these three gures the arc measure of ADC is equal to the degree measure of AOC by Theorem 37 But 1 we also know in these three cases that ABC ELAOC by part a of Theorem 36 so we conclude that ABC ADC for these three cases In Figure 39 the arc ADC is the part of the circle outside AOC so in this gure the arc measure of ADC is equal to 360 7 AOC by Theorem 37 BF 6 and algebra 1 But we also know that ABC 180 7 EAAOC in this case by part b of Theorem 1 36 so we conclude that ABC EADC in this case also QED Figure 36 Figure 37 B D O A C I B Flgure 38 Figure 39 58 Note The reason why Theorem 36 is more complicated than Theorem 38 is that the angles ABC and AOC don t always cut off the same arc as you can see from Figures 36 37 38 and 39 Since the relation between ABC and the arc it cuts off is simple this forces the relation between ABC and AOC to be complicated Next consider the situation where we have four points A B C and D on a circle and we are interested in the angles ABC and ADC The arcs cut off by ABC and ADC will always have the same endpoints namely A and C but they won t always be the same arc see Figure 40 in the rst picture the two angles do cut off the same arc but in the second picture they dont Figure 40 Theorem 39 Let A B C and D be points on a circle and consider the angles ABC and ADC a If ABC and ADC cut o the same arc then ABC ADC b If ABC and ADC do not cut o the same arc then ABC 180 7 ADC Proof Part a is immediate from Theorem 38 For part b see the second picture in Figure 40 Observe that the arc cut off by ABC is ADC and the arc cut off by ADC is ABC By BF 6 and algebra arc ADC is 360 minus arc ABC Now we have 1 ABC EADC by Theorem 38 360 7ABC by m 180 7 ABC by algebra 180 7 ADC by Theorem 38 59 which is what we were to show QED Theorem 39 is often useful7 and in particular it can be used to prove the following Theorem 40 Let A B C and D be points on a circle and suppose that the lines AB and CD meet at a point P Then PA PB PC PD4 There are two cases7 which are illustrated in Figures 41 and 42 The proofs of the two cases are homework problems Figure 41 Figure 42 64 Tangents to Circles De nition of tangent line A line is tangent to a circle ltgt it intersects the circle in exactly one point 60 Theorem 41 LctC be a circle with center 0 let A be a point on the circle and let in be a line through A Then in is tangent to the circle ltgt m is perpendicular to 0A Proof For the lt direction see Figure 43 Choose a point of in other than A and call it B Since in is perpendicular to 0A we can apply Theorem 8 to get 0132 0A2 A32 Since AB2 gt 0 this tells us that OB2 gt 0A2 and this implies OB gt GA But 0A is the radius of the circle so B cannot be on the circle by de nition of circle Therefore A is the only point of m which is on the circle so in is tangent to the circle by de nition of tangent The gt direction is a homework problem QED Figure 43 Next we have an analog of Theorem 38 when one of the sides of the angle is tangent to the circle Arc cut off by an angle tangent to the circle HA and B are points on a circle and C is a point on the tangent line at B then the arc cut off by ABC is the part of the circle inside ABC see Figure 44 61 Figure 44 Theorem 42 Let A and B be points on a circle let 0 be a point on the tangent line at B and let ADB be the arc cut o by ABCF Then ABC arc ADB Proof There are three cases i O is on ray BA ii 0 is inside ABC iii 0 is outside ZABC For case see Figure 45 Here are ADB is a semicircle7 so it measures 180 7 and LABC is a right angle by Theorem 41 so ABC is half of arc ADB in this case Figure 45 62 For case ii see Figure 46 Draw line OB and give the intersection of this line with the circle the name E Then LABC 4142 byBF6 1 1 Earc AFE Earc EDB by Theorem 38 and Case arc AFE are EDB algebra 1 Earc ADB by BF 6 applied to arcs D Figure 46 For Case iii see Figure 47 Draw line OB and give the intersection of this line with the circle the name E Then LABC QED EBC 7 1 by BF 6 and algebra 1 Earc EDB 7 arc EFA by Case and Theorem 38 1 arc EDB 7 are EFA algebra 1 arc ADB by BF 6 applied to arcs and algebra 63 Figure 47 7 Euclid s Elements Introduction Euclid s book was written around 300 BC It was the nishing step in a developmental process that went back about 300 years but unfortunately we have very little evidence of what the earlier stages of Greek geometry looked like Euclid s text was so successful that the earlier versions ceased to be copied by hand which was the only way for books to be preserved in those days Greek research in geometry continued for several hundred more years but Euclid s text was established as the de nitive account of the foundations of geometry and it retained this position until the early 19th century What was special about Greek geometry Before the time of the Greeks there were already ancient peoples who knew many geometrical facts for instance the Babylonians and the Egyptians and since their time there have been other cul tures such as the Hindus the Chinese and the Mayans who rediscovered various facts including the Pythagorean theorem independently The new feature of Greek geometry which had never existed before and which has never been discovered independently by other cultures is the idea of proof It is also true that the Greeks had factual knowledge about geometry which was vastly more extensive and sophisticated that of any other culture which pursued the subject These two features of the Greek achievement were not unrelated as we shall see 64 What is proof Brie y7 proof means explaining why a given fact is true by showing how it follows logically from simpler facts When we do this it is important to avoid circular argument7 so we need to be sure that the simpler facts can themselves be explained without using the fact we are proving Here is a simple example of a circular argument There s no school today How do you know My brother told me How does he know I told him This in turn means that we have to have some way of keeping track of what has been used to prove what Euclid does this by numbering the propositions and never using a later one to prove an earlier one What are postulates and common notions It is important to realize that not every statement can be reduced to simpler pieces7 since this would involve an in nite process of explaining each statement in terms of simpler ones7 and then explaining those in terms of even simpler ones7 and so on forever We have to start somewhere7 with statements whose truth is self evident ln Euclid7 these statements are the postulates and the axioms axioma is the Greek word which Heath translates as common notions The difference between an axiom and a postulate in Euclid is that an axiom is a statement which could apply to other parts of mathematics besides geometry and a postulate is a statement speci cally about geometry To sum up a theorem or proposition is a true fact which can be explained in terms of simpler facts A postulate or axiom is a true fact which cannot be explained in terms of simpler facts Incidentally7 a de nition unlike a postulate or axiom is not a statement about reality It is an agreement to use a certain word or short phrase to stand for a longer phrase For example7 we use the word parallelogram to stand for a four sided gure with two pairs of parallel sides Euclid s system has two very remarkable features which may not be obvious at rst glance The rst is that he has built all of geometry into a single uni ed system7 where each statement follows ultimately from a few simple postulates and axioms The second is the small number of the postulates and axioms and their simplicity This can be illustrated with an analogy from chemistry every chemical compound7 no matter how complicated7 can be broken down into the 92 naturally occurring elements What Euclid showed was that in geometry the situation is even simpler instead of 92 basic ingredients only 10 are needed The relation between proof and discovery It is important to realize that proving and discovering are not unrelated activities by sharpening their ability to explain a given fact in terms of known ones the Greeks also developed the skill of using 65 thought to pass from known facts to new discoveries The reason for this is that when the facts already known are organized and explained logically they can be used as a new basis for the imagination and a strengthened imagination can explore new areas that would otherwise be out of reach You have experienced something like this already in doing the homework problems Why does Euclid prove things that are obvious For example Proposition 5 is obvious from sight and yet Euclid gives a rather elaborate proof of it The question of why he does this is a very important one which comes up even in high school mathematics and you should decide for yourself what you think about it it is worth knowing that even Isaac Newton one of the greatest of mathematicians had questions on this subject Here are some relevant observations o If Euclid didn t prove the facts in Book 1 which are obvious from sight then he would have to add them to the list of postulates because a true fact about geometry which cant be explained in terms of simpler facts is a postulate and this means that instead of having a simple list of 5 postulates we would have about 15 o For the Greeks it was important not to rely on perceptions such as sight to establish the truth of things because everyone knows that perceptions can be mistaken as in optical illusions and the Greeks wanted to discover truths about mathematics and other things that didn t suffer from human limitations but were eternally and universally valid 9 The exercise of proving something which is obvious from sight is one of the best ways to sharpen your sensitivity to subtle logical issues This brings me to the topic of Cleverness and subtlety Cleverness and subtlety are two different things It is a little hard to describe each of them precisely but here is an attempt Some ways in which cleverness might be used in solving a homework problem are nding the right pair of congruent or similar triangles or drawing in the right extra line Subtlety often seems like a logical trick the main example we have seen so far is the indirect approach used in Theorems 16 17 22 23 and 26 Proof by contradiction is also an example of subtlety The main point I want to make right now is that subtlety isn t just for logical tricksters but is a tool for solving problems which is just as important as cleverness So here is one reason why Euclid proves things that are obvious it improves our ability to be subtle and this in turn improves our ability to solve problems Flaws in Euclid Since Euclid was a pioneer in organizing geometry as an axiomatic system it is not surprising that he made some mistakes more precisely omissions 66 mainly in the form of missing axioms I will point some of these out when we come to them and you will notice others It is interesting that almost none of these aws was observed until the nineteenth century when the discovery of non Euclidean geometry caused mathematicians to take a very close look at Euclid For the moment I want to observe that Euclid got it 90795 right and that nineteenth century mathematicians completed his work so that there are no aws at all any more Learning from the historical process You might ask why since Euclid s treatment is not perfect we use his version instead of some other Here are some reasons 9 Euclid s version is a good place to start since it is closer to your own point of view than a more perfected version would be He focuses on issues that you can learn from and does not obscure them with more technical issues that are necessary for complete perfection o A related point is that Euclid s language is closer to ordinary language than a perfected treatment would be 9 You can sharpen your own understanding by learning to spot the missing steps The in uence of Euclid on modern mathematics Until the nineteenth century geometry was the only part of mathematics which was organized on an axiomatic basis In the nineteenth century mathematicians began to encounter problems in volving Fourier series and other parts of calculus which could not be answered con dently without the kind of close analysis provided by an axiomatic approach They therefore undertook the project of reorganizing calculus on the Euclidean model This led to the formal de nition of limit and eventually to a completely axiomatic and rigorous approach to calculus In the twentieth century all of math ematics has been organized on an axiomatic basis ultimately resting on the axioms of set theory 8 Comments on the de nitions postulates common notions and propositions of Euclid Book I The de nitions Many of the de nitions such as those for point and line are not used at all in the proofs A closer examination of some of these de nitions shows that they are not really de nitions in the mathematical sense they dont describe the thing in terms of simpler things as the de nition of parallelogram does The de nitions that are used later are 10 15 16 20 22 and 23 67 Note that in de nition 20 an equilateral triangle is not considered to be isosceles and in de nition 22 a square is not considered to be an oblong what does Euclid mean by oblong This is different from the modern practice It is not uncommon for different versions of a subject to use somewhat different de nitions this is allowable because de nitions are not statements about reality but are convenient abbreviations which the author and his or her readers agree on Over the course of time it has been found to be more convenient to consider an equilateral triangle as a special kind of isosceles triangle rather than to require that an isosceles triangle has two but not three equal sides which is what Euclid does Within a given treatment of geometry the de nitions can t be changed the author has to use each word with the same meaning all the time Postulate 1 This is our BF 7 except that Euclid doesn7t say explicitly that there is only one line through two given points Postulate 2 This is our BF 9 Postulate 3 We didn t assume this as a Basic Fact but Euclid will use it to prove some of the Basic Facts that we did assume It is important to realize that this Postulate allows us to do Circle by center and point 7 but not Circle by center and radius 7 This is the opposite of what you might eXpect from the way it is worded I will eXplain this in class Postulate 4 This really is a statement about reality and not a de nition as I will eXplain in class Postulate 5 This postulate doesnt seem nearly as self evident as the others so for about 2300 years people tried to prove it Eventually they discovered that it cannot be proved from the other postulates and common notionsisurprisingly this led to a new subject non Euclidean geometry Common Notion 1 This is transitivity Common Notions 2 and 3 These are included in what we have been calling alge bra77 Euclid s goal here is to state in advance what kinds of algebra he will allow although he does sometimes use things not included in Common Notions 2 and 3 as we shall see Proposition 1 Frequently asked questions What does circle BCD77 mean in line 8 Answer Euclid names a circle by giving three points on it Sometimes this means that he has to create a point whose only purpose is to name the circle it7s onithat7s what he s done here with point D He has to give three points instead of just two because there is more than one 68 circle going through two given points but only one circle going through three given points What does Post 1 stand for in line 14 Answer Postulate 1 Incidentally all of the reasons given in brackets were supplied by the modern editor TL Heath and were not in Euclid s original manuscript On quizzes and exams you are responsible for giving all ofthe reasons including those that Heath forgets to give using only facts from Euclid and not from the course notes There is one exception you don t have to give citations for Postulate 1 since it occurs so often and is so elementary What does CN 1 stand for in line 22 Answer Common Notion 1 Like all construction proofs this one has two parts a recipe for the construction lines 1714 and a veri cation that the recipe does what it is supposed to do lines 15726 Notice that the proof ends with a summary Therefore the triangle ABC is equi lateral Also notice the very last line what it was required to do ithis is how Euclid ends a construction proof other proofs end with what it was required to show Proposition 2 Frequently asked questions What does 11 mean in line 10 Answer Book 1 Proposition 1 What does circle OCH mean in line 15 Answer Here Euclid is being a bit sloppy in his explanation What he really means is that we draw the circle that has center B and goes through 0 then we nd the intersection of this circle with the extended line DF and call this intersection point G The point H plays no role except to be part of the name of the circle What does circle GKL mean in line 16 Answer Here we draw the circle with center D that goes through G nd the intersection ofthis circle with the extended line DE and call this intersection L The point K plays no role except to be part of the name of this circle The purpose of this proof is to show how one can carry out Circle by center and radius using only Circle by center and point This is a construction proof so it has a recipe and a veri cation that the recipe works The recipe is given in lines 1717 and the veri cation is lines 18732 Proposition 3 This includes part of our BF 8 It says that on a given ray there is a point at a given distance from the endpoint but not that there is only one such point 69 Proposition 4 This is SAS We have assumed it as BF 27 but Euclid gives a proofof it Frequently asked question What does Therefore etc mean in line 44 Answer the editor has gotten tired of writing out the summary at the end of the proof7 so he uses this for an abbreviation see his note in the middle of page 249 Compare this to the last two lines in the proofs of Propositions 17 2 and 3 It is very important to note here that the triangle will be equal to the triangle in line 4 does not mean the triangles are congruent Euclid has no term corre sponding to our term congruent so whenever he wants to say that two triangles match up completely he has to mention all siX parts7 which gets tedious When he says that two triangles are equal he is de nitely not saying that they match up in all siX parts7 he is only saying that they have equal area This can be seen most clearly by looking at the statement and picture for Proposition 37 Proposition 5 This is one direction of our Theorem 5 Notice that we used BF 10 and BF 1 in our proof7 and that Euclid doesn7t yet have these tools eventually he proves BF 1 and BF 10 as Propositions 8 and 10 Frequently asked question What is happening in lines 15725 Answer Euclid is quoting Proposition 4 SAS Lines 15719 show that the hypotheses of Proposition 4 are satis ed in his situation7 and lines 20725 spell out what the conclusion of Proposition 4 says in his situation Note that he quotes Proposition 4 word for word instead of just referring to itipresumably this is to emphasize the point that he is specializing a general statement to a particular situation and that the hypotheses must match up exactly You should ignore the last part of the statement and7 if the equal straight lines be produced further7 the angles under the base will be equal to one another and its proof lines 46 47 Note that the proof ends with QED This stands for the Latin phrase Quod erat demonstrandum which is a translation of the Greek phrase which means what was to be shown The editor ends construction proofs from now on with QEF7 which stands for quod erat faciendum meaning what was to be done see his note about line 48 Proposition 6 This is the other direction of our Theorem 5 Notice that in line 16 the phrase the triangle BBC will be equal to the triangle ACB means that these triangles will have the same area Although in this case it is also true that the triangles will be congruent7 that is not the point that Euclid is drawing our attention to here The point of his argument is that if a triangle is contained inside another then its area must be smaller than that of the triangle it 70 is contained in7 so the areas ofthe two triangles cannot be equal7 the lesser to the greater This is our rst example of a proof by contradiction Proposition 7 This proposition is only used to prove Proposition 87 and will never be used again The modern name for a proposition which is used only as a tool to prove something else is lemmaf7 It is easier to follow the proof ofthis proposition if you use numbers instead of three letter names to label some of the angles Notice that in line 21 Euclid says that the angle CDB is much greater 7 than the angle DOB What is happening here is that Euclid is using the fact that if one thing is greater than a second7 and the second is greater than a third7 then the rst is greater in fact much greater than the third This sounds similar to Common Notion 1 but in fact it is a missing Common Notion that should have been added to the original list Proposition 8 This is SSS We have assumed this7 as BE 17 but Euclid proves it Propositions 9712 These are the usual constructions that are taught in high school Notice that as usual for construction proofs each of these proofs consists of a recipe followed by an explanation of why the recipe works Proposition 9 is our BF 117 Proposition 10 is our BF 107 and Propositions 11 and 12 are our BE 12 Propositions 13 and 14 These are converses of each other7 and Euclid uses Propo sition 137 plus proof by contradiction7 to prove Proposition 14 We will see this pattern7 of a statement being used to prove its converse by means of proof by con tradiction7 several times But it isn t always possible to prove the converse this way7 as we will see in discussing Propositions 27 and 29 Proposition 14 will be used later to prove that certain lines are straight Proposition 16 Frequently asked questions Why doesn7t Euclid just use the fact that the angles of a triangle add up to 180 Answer Because it isn t yet availableithe fact that the angles of a triangle add up to 180 is Proposition 32 OK7 then why doesnt Euclid just wait to prove Proposition 16 until after he has proved Proposition 32 Answer He can t do this because it would be a circular argumentl One of the steps in proving Proposition 32 is Proposition 31 this is the fact that it is possible to construct a parallel line through a given point7 and Proposition 31 in turn uses Proposition 277 and Proposition 27 uses Proposition 16 There is no way to prove Proposition 32 without proving Proposition 16 rst It s true that we proved it in the course notes without using Proposition 167 but 71 that s because we were assuming BF 13 Euclid proves BF 137 and to prove it he uses Proposition 16 This illustrates a basic difference between axiomatic systems and lists of facts if we think of Proposition 16 purely as a fact then it is included in Proposition 32 and therefore not very important But if we think about how to prove that Proposition 32 is true then Proposition 16 is an indispensable tool One more comment that is of some interest is that Proposition 16 since it doesnt make use of Postulate 5 is true in both Euclidean and non Euclidean geometry7 whereas Proposition 32 is not true in non Euclidean geometry Another frequently asked question in line 107 how do we know where F is supposed to be Answer Euclid is cutting corners a bit here and should have said it more clearly What he should have said is let BE be extended Postulate 2 and let a point F be marked on the extension with EF BE Proposition 3 7 This is the sort of thing that modern mathematicians are always very careful to spell out7 which is one reason why modern mathematical writing often seems cluttered Finally7 notice that the last part of the argument lines 27729 on page 280 isn t exactly similar to the rst part You should draw the pictures carefully to see what the difference is Propositions 18 and 19 This is another example where the statements are converses of each other and the second is proved from the rst using proof by contradiction Proposition 20 Read the editor s note about the Epicureans on page 287 What the Epicureans were asking was why do we prove things that are obvious 7 The Epicureans were intelligent people and we shouldnt be surprised ifour own students ask the same question I have already given my own answer to this earlier in these notes7 but at this point I can add to it you can probably see now that one reason we prove things that are obvious is that the proofs are often exciting examples of the art of proof Propositions 24 and 25 This is yet another example where the second proposition is the converse of the rst and is proved from it using proof by contradiction It is easier to follow the proof of Proposition 24 if you use numbers instead of three letter names to label some of the angles Proposition 26 This has two parts the rst is ASA proved on page 302 and the second is AAS proved on page 303 Frequently asked questions Why doesn7t Euclid just use ASA to prove AAS like we usually do Answer because in order to do this you need Proposition 32 which he doesn t yet have 72 OK7 then why doesnt Euclid delay Proposition 26 until after Proposition 32 Answer this is a harder question to answer He could have done that7 but we think that the reason he didn t was because he wanted to prove as many things as possible without using Postulate 5 Does this mean there s something wrong with Postulate 5 Answer No This is another place where the difference between an axiomatic system and a list of facts becomes important No one doubted that Postulate 5 was a true factithe question was whether it was really a postulate a true fact which cannot be proved or a proposition a true fact which can be proved Euclid clearly felt that it was a postulate7 but he was also willing to give people a chance to try to prove it if they wanted to and this seems to be why he showed how to prove Proposition 26 without it now Proposition 26 could be used in trying to prove Postulate 5 without creating a circular argument As a matter of historical fact7 for the next 2300 years people made many attempts to prove Postulate 57 all of which turned out to be incorrect We now know that it is impossible to prove Postulate 57 and this impossibility is closely related to the possibility of non Euclidean geometry Proposition 27 This is the second part of our Theorem 2a Proposition 28 This is the second part of our BF 5 and Theorem 2b Proposition 29 This is the rst parts of our BF 57 Theorem 2a and Theorem 2b Proposition 29 is the converse of Proposition 277 but in this case Euclid does not follow the pattern of proving the converse from the original statement by contra diction He could not follow the pattern if he wanted to because it is impossible try it for yourself lnstead7 he has to introduce a new ingredient that he hasn t used before7 namely Postulate 5 see line 24 of the proof Proposition 30 This is our BF 14 Frequently asked question Isn t this just transitivity Answer Transitivity is Common Notion 1 things which are equal to the same thing are also equal to one another In our case we are not asserting that the lines are equal77 but rather that they are parallel 7 which is different Of course7 Common Notion 1 is used in the proof line 18 because we can show that two lines are parallel by showing that certain angles are equal This proof is a very rare example of Euclid making a logical mistake in line 5 he says For let the straight line GK fall upon them77 but it isn t possible to create such a line without assuming Proposition 30 1711 show in class how to x the proof Proposition 31 This is our BF 13 We assume BF 13 and Euclid proves it 73 Proposition 34 Notice that the word parallelogram or rather parallelogrammic area is used here without having been de ned Euclid seems to assume that we know that the de nition of parallelogram is a four sided gure with two pairs of parallel sides It is a mystery why he didn t include this de nition in his list of de nitions Proposition 35 This is related to the fact which Euclid does not use or even mention that the area of a parallelogram is base times height In the case we are looking at the parallelograms have the same base7 and the fact that they are in the same parallels forces them to have the same height Frequently asked question Why doesn7t he just say that Answer The Greeks had recently discovered the existence of irrational numbers7 and because of this they regarded statements about measurement as uncertain and untrustworthy from a scienti c point of view Because of that7 Euclid never talks about the length of a segment The Greeks certainly knew our usual formulas for calculating area but they thought of these formulas as practical rules that didn t belong in a scienti c account Proposition 36 The editor makes a truly awful mistake here he gives Proposition 34 as the reason for the statement Therefore EBCH is a parallelogram He should have said that this is because of the missing de nition of parallelogram Proposition 37 This proof uses another missing common notion the halves of equal things are equal to one another Proposition 42 This is the rst step in proving Proposition 44 It is a good exercise to carry out the recipe in the proof of Proposition 42 on Geometer7s Sketchpad Proposition 43 Very frequently asked question What does EH mean in line 4 and elsewhere Answer Euclid has suddenly introduced this as a convenient way of abbreviating EKHA as the name of a parallelogram Similarly7 he refers to the parallelograms FCCHK7 BGKE7 and KFDH as FG7 BK7 and KD I don t know why he has decided to do this Proposition 44 This is a very powerful construction To give an idea of how powerful it is7 let us observe that we can use it to divide the length of one segment by the length of another using ruler and compass alone To do this7 we let the given straight line be the second segment7 the given rectilineal angle be a right angle7 and the given triangle be a right triangle with height two and base equal to the rst segment Then the side of the parallelogram actually a rectangle in this case which is constructed by Proposition 44 will be the quotient of the rst segment by the second segment 74 Again7 it is a good exercise to carry out the recipe in this proof using Geometer7s Sketchpad for this you need to have a script that does the recipe in Proposition 42 I recommend that you read the note which starts at the bottom of page 342 about the godlike men of old Proposition 45 This proposition allows us to transform any given polygonal area into a rectangle with the same area7 using ruler and compass alone Proposition 47 This is what most of Book I has been leading up to Euclid s proof of the Pythagorean theorem The ancient commentators seem to agree that this proof is due to Euclid himself7 which leads to the question of what Pythagoras s original proofwas The answer seems to be that Pythagoras s original proofwas the same as the one in the course notes using similar triangles Between the time of Pythagoras and Euclid the Greeks discovered irrational numbers7 which made the whole question of what is meant by proportional sides much more complicatedi this is why Euclid doesn t mention similar triangles anywhere in Book I Eventually Euclid gives a full treatment of similarity and proves our BF 4 in Book VI7 but he didn t want to delay the Pythagorean theorem until then so he gives a more elementary proof now Notice that in the diagram on page 349 the lines AD7 FC and BK appear to be concurrentithey are in fact concurrent and this can be proved by Theorem 35 Proposition 48 This is the converse to the Pythagorean Theorem Euclid s proof uses the fact that if two squares have equal area then their sides are equal Notice that he doesn7t give any justi cation for this7 which is a mistake on his part maybe he was tired after the elaborate proof of Proposition 47 The missing statement sounds like a common notion similar to equals added to equals are equal but in fact it can be proved7 and the editor gives a proof at the bottom of page 348 This is a good example of how hard it can be to tell whether a simple looking statement can be proved by using even simpler statements 75

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