Cevians, Symmedians, and Excircles. MA 341 Topics in Geometry Lecture 16

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1 Cevians, Symmedians, and Excircles MA 341 Topics in Geometry Lecture 16

2 Cevian A cevian is a line segment which joins a vertex of a triangle with a point on the opposite side (or its extension). B cevian A D C 05-Oct-2011 MA

3 Cevian Triangle & Circle Pick P in the interior of ABC Draw cevians from each vertex through h P to the opposite side Gives set of three intersecting cevians AA, BB, and CC with respect to that point. The triangle A B C ABC is known as the cevian triangle of ABC with respect to P Circumcircle of A B C is known as the evian circle with respect to P. 05-Oct-2011 MA

4 Cevian circle Cevian triangle 05-Oct-2011 MA

5 Cevians In ABC examples of cevians are: medians cevian point = G perpendicular bisectors cevian point = O angle bisectors cevian point = I (incenter) altitudes cevian point = H Ceva s Theorem deals with concurrence of any set of cevians. 05-Oct-2011 MA

6 Gergonne Point In ABC find the incircle and points of tangency of incirclei with ih sides of ABC. Known as contact triangle 05-Oct-2011 MA

7 Gergonne Point These cevians are concurrent! Why? Recall that AE=AF, BD=BF, and CD=CE Ge 05-Oct-2011 MA

8 Gergonne Point The point is called the Gergonne point, Ge. Ge 05-Oct-2011 MA

9 Gergonne Point Draw lines parallel to sides of contact triangle through Ge. 05-Oct-2011 MA

10 Gergonne Point Six points are concyclic!! Called the Adams Circle 05-Oct-2011 MA

11 Gergonne Point Center of Adams circle = incenter of ABC 05-Oct-2011 MA

12 Isogonal Conjugates Two lines AB and AC through vertex A are said to be isogonal if one is the reflection of the other through the angle bisector. 05-Oct-2011 MA

13 Isogonal Conjugates If lines through A, B, and C are concurrent at P, then the isogonal lines are concurrent at Q. Points P and Q are isogonal conjugates. 05-Oct-2011 MA

14 Symmedians In ABC, the symmedian AS a is a cevian through vertex A (S a BC) isogonally conjugate to the median AM a, M a being the midpoint of BC. The other two symmedians BS b and CS c are defined similarly. 05-Oct-2011 MA

15 Symmedians The three symmedians AS a, BS b and CS c concur in a point commonly denoted K and variably known as either the symmedian point or the Lemoine point 05-Oct-2011 MA

16 Symmedian of Right Triangle The symmedian point K of a right triangle is the midpoint of the altitude to the hypotenuse. A K M b B D C 05-Oct-2011 MA

17 Proportions of the Symmedian Draw the cevian from vertex A, through the symmedian point, to the opposite side of the triangle, meeting BC at S a. Then c b BS 2 a c 2 CS b a a 05-Oct-2011 MA

18 Length of the Symmedian Draw the cevian from vertex C, through the symmedian point, to the opposite side of the triangle. Then this segment has length Likewise ab 2a 2b c CS c 2 2 a b bc 2b 2c a AS a 2 2 b c ac 2a 2c b BS b 2 2 a c Oct-2011 MA

19 Excircles In several versions of geometry triangles are defined in terms of lines not segments. A B C 05-Oct-2011 MA

20 Excircles Do these sets of three lines define circles? Known as tritangent circles A B C 05-Oct-2011 MA

21 Excircles I C B r c A I r b I B C I A r a 05-Oct-2011 MA

22 Construction of Excircles 05-Oct-2011 MA

23 Extend the sides 05-Oct-2011 MA

24 Bisect exterior angle at A 05-Oct-2011 MA

25 Bisect exterior angle at B 05-Oct-2011 MA

26 Find intersection I c 05-Oct-2011 MA

27 Drop perpendicular to AB I c 05-Oct-2011 MA

28 Find point of intersection with AB I c 05-Oct-2011 MA

29 Construct circle centered at I c I c r c 05-Oct-2011 MA

30 05-Oct-2011 MA

31 Excircles The I a, I b, and I c are called excenters. r a, r b, r c are called exradii 05-Oct-2011 MA

32 Excircles Theorem: The length of the tangent from a vertex to the opposite exscribed circle equals the semiperimeter, s. CP = s 05-Oct-2011 MA

33 Excircles 1. CQ = CP 2. AP = AY 3. CP = CA+AP = CA+AY 4. CQ= BC+BY 5. CP + CQ = AC + AY + BY + BC 6. 2CP = AB + BC + AC = 2s 7. CP = s 05-Oct-2011 MA

34 Exradii 1. CP I C P 2. tan(c/2)=r C /s 3. Use Law of Tangents I c C (s a)(s b) s(s a)(s b) r stan s c 2 s(s c) s c 05-Oct-2011 MA

35 Exradii Likewise r a r b r c s(s b)(s c) s a s(s a)(s c) s b s(s a)(s b) s c 05-Oct-2011 MA

36 Excircles Theorem: For any triangle ABC r r r r a b c 05-Oct-2011 MA

37 Excircles s a s b s c r r r s(s b)(s c) s(s a)(s c) s(s a)(s b) a b c s a s b s c s(s a)(s b)(s c) s(s a)(s b)(s c) s(s a)(s b)(s c) 3s (a b c) s(s a)(s b)(s c) s s(s a)(s b)(s c) s K 1 r 05-Oct-2011 MA

38 Nagel Point In ABC find the excircles and points of tangency of the excircles with ih sides of ABC. 05-Oct-2011 MA

39 Nagel Point These cevians are concurrent! 05-Oct-2011 MA

40 Nagel Point Point is known as the Nagel point 05-Oct-2011 MA

41 Mittenpunkt Point The mittenpunkt of ABC is the symmedian point of the excentral triangle ( I a I b I c formed from centers of excircles) 05-Oct-2011 MA

42 Mittenpunkt Point The mittenpunkt of ABC is the point of intersection of the lines from the excenters through midpoints of corresponding sides 05-Oct-2011 MA

43 Spieker Point The Spieker center is center of Spieker circle, i.e.,., the incenter of the medial triangle of the original triangle. 05-Oct-2011 MA

44 Special Segments Gergonne point, centroid and mittenpunkt are collinear GGe =2 GM 05-Oct-2011 MA

45 Special Segments Mittenpunkt, Spieker center and orthocenter are collinear 05-Oct-2011 MA

46 Special Segments Mittenpunkt, incenter and symmedian point K are collinear with distance ratio IM 2(a +b + c ) = 2 MK (a +b +c) 05-Oct-2011 MA

47 Nagel Line The Nagel line is the line on which the incenter, triangle centroid, Spieker center Sp, and Nagel point Na lie. GNa =2 IG 05-Oct-2011 MA

48 Various Centers 05-Oct-2011 MA

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