Cevians, Symmedians, and Excircles. Cevian. Cevian Triangle & Circle 10/5/2011. MA 341 Topics in Geometry Lecture 16
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1 Cevians, Symmedians, and MA 341 Topics in Geometry Lecture 16 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 Cevian Triangle & Circle Pick P in the interior of ABC Draw cevians from each vertex through 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 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
2 Cevian circle Cevian triangle 05-Oct-2011 MA 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 Gergonne Point In ABC find the incircle and points of tangency of incircle with sides of ABC. Known as contact triangle 05-Oct-2011 MA
3 Gergonne Point These cevians are concurrent! Why? Recall that AE=AF, BD=BF, and CD=CE Ge 05-Oct-2011 MA Gergonne Point The point is called the Gergonne point, Ge. Ge 05-Oct-2011 MA Gergonne Point Draw lines parallel to sides of contact triangle through Ge. 05-Oct-2011 MA
4 Gergonne Point Six points are concyclic!! Called the Adams Circle 05-Oct-2011 MA Gergonne Point Center of Adams circle = incenter of ABC 05-Oct-2011 MA 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
5 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 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 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
6 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 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 CS a a c b 2 2 a 05-Oct-2011 MA 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 ab 2a 2b c CS c 2 2 a b Likewise bc 2b 2c a AS a 2 2 b c ac 2a 2c b BS b 2 2 a c 05-Oct-2011 MA
7 In several versions of geometry triangles are defined in terms of lines not segments. A B C 05-Oct-2011 MA Do these sets of three lines define circles? Known as tritangent circles A B C 05-Oct-2011 MA I C B r c A I I B r b C I A r a 05-Oct-2011 MA
8 Construction of 05-Oct-2011 MA Extend the sides 05-Oct-2011 MA Bisect exterior angle at A 05-Oct-2011 MA
9 Bisect exterior angle at B 05-Oct-2011 MA Find intersection I c 05-Oct-2011 MA Drop perpendicular to AB I c 05-Oct-2011 MA
10 Find point of intersection with AB I c 05-Oct-2011 MA Construct circle centered at I c I c r c 05-Oct-2011 MA Oct-2011 MA
11 The I a, I b, and I c are called excenters. r a, r b, r c are called exradii 05-Oct-2011 MA Theorem: The length of the tangent from a vertex to the opposite exscribed circle equals the semiperimeter, s. CP = s 05-Oct-2011 MA 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
12 1. CP I C P 2. tan(c/2)=r C /s 3. Use Law of Tangents Exradii 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 Likewise Exradii 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 Theorem: For any triangle ABC r r r r a b c 05-Oct-2011 MA
13 1 1 1 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(s a)(s b)(s c) K 1 r 05-Oct-2011 MA Nagel Point In ABC find the excircles and points of tangency of the excircles with sides of ABC. 05-Oct-2011 MA Nagel Point These cevians are concurrent! 05-Oct-2011 MA
14 Nagel Point Point is known as the Nagel point 05-Oct-2011 MA 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 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
15 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 Special Segments Gergonne point, centroid and mittenpunkt are collinear GGe =2 GM 05-Oct-2011 MA Special Segments Mittenpunkt, Spieker center and orthocenter are collinear 05-Oct-2011 MA
16 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 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 Various Centers 05-Oct-2011 MA
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