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# Fund Concepts of Geometry MATH 3181

UNCC

GPA 3.73

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This 5 page Class Notes was uploaded by Mrs. Dangelo Fahey on Sunday October 25, 2015. The Class Notes belongs to MATH 3181 at University of North Carolina - Charlotte taught by Staff in Fall. Since its upload, it has received 63 views. For similar materials see /class/228920/math-3181-university-of-north-carolina-charlotte in Mathematics (M) at University of North Carolina - Charlotte.

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Date Created: 10/25/15

Newer Results in Euclidean Geometry Proposition 1 Let AABC be a triangle and let D E be the midpoints of AB and AC respectively Then the line DE is parallel to the base Bc and congruent to one half of it In other words ifF is the midpoint ofBC then DE 5 BF Proof Let D be the midpoint of AB and draw lines through D parallel to AC and BC Let them meet the opposite sides in points E and F Since DE is parallel to BC the angles ADBF E ADE are congruent Similarly since AC is parallel to DF the angles ADAE 2 BDF are congruent Now AD 5 DB and the angles of the triangles AADE and ADBF are congruent so by ASA the two triangles are congruent Therefore AE 2 DF and DE E BF Look at the parallelogram DE F C Since it is a parallelogram opposite sides are congruent Thus DF 2 CE and DE 2 CF Thus E and F are the midpoints of the sides AC and BC respectively Thus E E and the line DE is the line DE Thus DE is parallel to BC as claimed Furthermore DEDE BF BFBC Corollary Let AABC be a triangle and let D E and F be the midpoints of the sides Then the sides of the triangle ADEF are parallel to the sides of AABC and the four triangles so formed are all congruent to one another We say atriangle AABC is congruent to the double of ADEF symbolically AABC 2 ZADEF if the two triangles are similar and the ratio of the sides is 21 Proposition 2 2 ASA Let AABC and AA B C be two triangles and assume that AABC E AA B C and zACB E ZA C B and that BC E 2B C Then AABC E 2AA B C Proof Let D E and F be the midpoints of the sides of AABC and draw the triangle ADEF From above we know that DE 2BC EB C Furthermore since De is parallel to BC the angles of AADE at D and E are congruent to the angles of AA B C at B and C Thus by ASA AADE E AA B C Since AABC is the double of AADE the result follows Similar theorems for SSS and SAS can be proven Proposition 3 The medians of a triangle meet in a single point called the centroid of the triangle Proof Let AABC be the triangle let D andE be the midpoints of AB and AC and construct DE Let the two medians BE and CD meet at a point G Since DE is parallel to BC we see that ADEGE ZCBG and AEDGE ABCG Also we A know that BC 5 2DE Thus by our previous proposition we see that ABGC E 2AEGD In particular it follows that BG 5 2GE Therefore G E is the point on the median BE that is 23 of the way D from B to E If we draw the medians BE and AF we find that they meet in a point that is also 23 of the way from B to C E Thus this must be the same point and all three medians meet in a single point G 5 Corollary The centroid G lies on each median 23 of the way from the vertex to the midpoint of the opposite side Proposition 4 The three altitudes of a triangle meet in a single point called the orthocenter of the triangle Proof Let AABC be the given triangle Construct lines through the vertices A B and C that are parallel to the opposite sides This will form a new triangle AA B C From the parallelograms BCAC and BCB A we nd that CA 5 BC 5 AB Thus A is the midpoint of B C and similarly for the other two sides of AA B C On the other hand the altitude AM of the triangle AABC is perpendicular to BC and hence is perpendicular to B C Thus the altitudes of AABC correspond to the perpendicular bisectors of the sides of AA B C From our previous result they meet in a single point How do these four points relate to one another Proposition 5 The Euler line In a triangle AABC let 0 be the circumcenter G the centroid and H the orthocenter Then 0 G and H lie on a line called the Euler line of the triangle and GH E 20G Proof For the proof let F be the midpoint of BC construct the median AF and let the line OG meet the altitude AM in a point H OF is perpendicular to the side BC since 0 is the circumcenter Thus OF is parallel to AM Hence ZGAH E ZGFO Also ZAGH 5 LEGO since they are vertical angles From the above Proposition about the medians we have that AGEZGF Thus by 2ASA we have that AAGH E ZAFGO Thus GH E 20G Thus the point H has the property of being the point on ray E so that GH EZOG By interchanging the roles ofA B and C it follows that H also lies on the other altitudes of AABC so H H is the orthocenter and our conclusions about 0 G and H begin collinear and GH 5 20G follow There is another point of interest associated with a triangle First we need some preliminary results to make our proof easier A cyclic quadrilateral is a set of four points A B C and D lying in that order on a circle together with the lines AB BC CD and DA joining them The lines AC and BD are the diagonals of the cyclic quadrilateral The importance of a cyclic quadrilateral comes from the relationships between the various angles Proposition 6 Let A B C anal D be four points in the plane with A anal B both on the same siale of the line CD Then A B C anal D lie on a circle ifanal only ifADAC E ADBC Proof If the points lie on a circle then the angles must be congruent since they subtend the same arc DC Now suppose that the two angles are congruent Construct the circle through A C and D and let it intersect the line BD at a point X 5 Since ADXC subtends the same arc as ADAC we have that X ADBC EADACEADXC This contradicts the Exterior Angle Theorem since ADXC is exterior to ACXB Thus X B and all four points lie on the circle Proposition 7 The NinePoint Circle Theorem In any triangle the midpoints of the three siales the feet of the three altitudes anal the midpoints of the three segments to the orthocenter all lie on a circle Proof Let AABC be our triangle Let M1 M2 and M3 be the midpoints ofthe sides F1 F2 and F3 the feet ofthe altitudes H the orthocenter and M4 M5 and M5 the midpoints of the segments joining H to the vertices We need to show that these nine points lie on one circle C We are going to use Proposition 1 several times First applied to AABC we see that M1M2 is parallel to BC In ABCH we see that M5M is parallel to BC and hence to M1M2 Now apply this proposition to AACH and AABH We get that M2M5 is parallel to AH and M1M5 is parallel to AH and hence to each other Now AH is perpendicular to BC altitude hence M 2M 5 is perpendicular to M1M2 and M6MS Therefore M1M2M5l5 is a rectangle Let P be the point of intersection of the diagonals of this rectangle P is equidistant from the points M1 M2 M 5 and M 5 Thus there is a circle centered at P containing those four points We only need to show that the other points lie on this circle M1F2M2M5 is a cyclic quadrilateral because the angles at M1 and F2 subtending the arc MZMS are right angles Since three points determine a circle this circle is our circle since it contains M1 M2 and M5 Hence F2 lies on the circle Using a similar argument to show M1F1F2M5 is a cyclic quadrilateral puts F2 on the circle Now shift perspective so that AC is regarded as the base of the triangle The same argument shows that M1M4F2M3 is a rectangle with the same center P Thus M4 and M3 lie on the circle Finally F3A1F1M3 is a cyclic quadrilateral as above so F3 lies on the circle 0 This point P is called the ninepoint center Proposition 8 The nine point center lies on the Euler line The nine point center is the midpoint of the Euler segment OH The centroid is one third of the way fromOtoH There are a number of interesting properties about the ninepoint circle 1 The ninepoint circle is the circumcircle of the medial triangle A 2 The ninepoint circle has radius one half that of the circumcircle 3 The ninepoint circle is the circumcircle of the triangle whose vertices are the midpoints of the segments joining AABC39s vertices to the orthocenter 4 The ninepoint circle passes through the points where AABC39s sides are cut by the lines that join AABC s vertices with its orthocenter The picture above shows AABC with its incenter I circumcenter O ninepoint center P centroid G and orthocenter H Note that the incenter does not lie on the Euler segment for this triangle It does lie on the Euler segment for some triangles and those can be characterized It does appear though that the incenter has been le out There are some other points of interest of a triangle These points help tie together the Euler segment and the incenter An excircle of a triangle AABC is a circle outside the triangle that is tangent to all three of the lines that extend the sides of the triangle There are three such circles each tangent to a side and the extensions of the other two sides If you connect the point of tangency to the opposite vertex these segments will intersect in a point lt called the Nagel point N One more point of interest is the center of the incircle for AABC39s medial triangle This circle is called the Spieker circle and its center is called the Spieker point Proposition 9 The Nagel segment is a line segment from the incenter I to the Nagel point N which contains the Spieker point S and the centroid G There is more about this Nagel segment and the Spieker circle 1 The Spieker circle is the incircle of AABC39s medial triangle 2 The Spieker circle has radius onehalf of AABC39s incircle 3 The Spieker circle is the incircle of the triangle whose vertices are the midpoints of the segments that join ABC s vertices with its Nagel point 4 The Spieker circle is tangent to the sides of AABC39s medial triangle where that triangle s sides are cut by the lines that join AABC39s vertices with its Nagel point Note the similarity to the ninepoint circle In addition we have the following Proposition 10 The Spieker point is the midpoint of the Nagel segment The centroid is one third of the way from the incenter to the Nagel point

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