More Incenter Properties (Inspired by Alan Yan)

I have noticed that our fellow member has been posting a few proof challenges regarding the incircle and excircle configurations, and I thought why not join him and spread more awesome properties as challenges. Perhaps we could eventually compile them into a wiki?

Let the incircle of ABCABC touch BC,AC,ABBC,AC,AB at D,E,FD,E,F, MM is the midpoint of BCBC. If AM∩EF=KAM\cap EF=K, prove that KD⊥BCKD\perp BC.

#Geometry

Note by Xuming Liang
5 years, 6 months ago

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Comments

@Alan Yan There are proofs of this property that involve harmonic division+poles and polar, as well as law of sines. Here I present a fairly simple synthetic solution:

Let II be the incenter and define DI∩EF=K′DI\cap EF=K', since ID⊥BCID\perp BC, it suffices to prove K=K′K=K'.

Through K′K' construct the line perpendicular to DK′DK' that intersects AB,ACAB,AC at Y,XY,X. Since IF⊥AB,IE⊥ACIF\perp AB, IE\perp AC, IK′YF,IK′EXIK'YF, IK'EX are cyclic quadrilateral. Hence ∠K′YI=∠EFI=∠FEI=∠K′XI\angle K'YI=\angle EFI=\angle FEI=\angle K'XI, which means IX=IYIX=IY. Since IK′⊥XYIK'\perp XY, so K′K' is the midpoint of XYXY. Because XY∥BCXY\parallel BC, A,K′,MA,K',M are collinear and thus K=AM∩EF=K′K=AM\cap EF=K'. Q.E.D

Xuming Liang - 5 years, 6 months ago

This solution is "not-so-elegant". I will try to find one when I have time.

The "easy to find points" are practically crying out barycentric coordinates. Letting A=(1,0,0),B=(0,1,0),C=(0,0,1)A = (1, 0, 0) , B = (0,1,0), C=(0,0,1), we have D=(0:s−c:s−b)E=(s−c:0:s−a)F=(s−b:s−a:0)M=(0:1:1)\begin{aligned} D & = (0 : s-c: s-b)\\ E & = (s-c:0:s-a) \\ F & =(s-b : s-a : 0)\\ M & = (0:1:1) \\ \end{aligned} We can find easily the line equations for AMAM and FEFE. AM:y=zFE:(a−s)x+(s−b)y+(s−c)z=0 \begin{aligned} AM&: y = z \\ FE&: (a-s)x + (s-b)y + (s-c)z = 0 \\ \end{aligned} Thus we solve these two equations to get that K=(a:s−a:s−a)K = (a : s-a : s-a) . It suffices to prove that KK lies on the line equation of IDID where II is the incenter. This is because ID⊥BCID \perp BC. Since I=(a:b:c)I = (a:b:c) we find that the line equation for IDID is (b−c)(s−a)ax+(s−b)y+(c−s)z=0.\frac{(b-c)(s-a)}{a}x + (s-b)y + (c-s)z = 0.

Plugging in KK we get (b−c)(s−a)+(s−b)(s−a)+(c−s)(s−a)=(s−a)[(b−c)+(s−b)+(c−s)]=0 (b-c)(s-a) + (s-b)(s-a) + (c-s)(s-a) = (s-a)[(b-c) + (s-b) + (c-s)] = 0 which is indeed zero. Thus, because Geometry is Algebra :P, we are done.

Alan Yan - 5 years, 6 months ago
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