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# 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 $$ABC$$ touch $$BC,AC,AB$$ at $$D,E,F$$, $$M$$ is the midpoint of $$BC$$. If $$AM\cap EF=K$$, prove that $$KD\perp BC$$.

Note by Xuming Liang
2 years, 1 month ago

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@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 $$I$$ be the incenter and define $$DI\cap EF=K'$$, since $$ID\perp BC$$, it suffices to prove $$K=K'$$.

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

- 2 years, 1 month 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 $$A = (1, 0, 0) , B = (0,1,0), C=(0,0,1)$$, we have \begin{align*} D & = (0 : s-c: s-b)\\ E & = (s-c:0:s-a) \\ F & =(s-b : s-a : 0)\\ M & = (0:1:1) \\ \end{align*} We can find easily the line equations for $$AM$$ and $$FE$$. \begin{align*} AM&: y = z \\ FE&: (a-s)x + (s-b)y + (s-c)z = 0 \\ \end{align*} Thus we solve these two equations to get that $$K = (a : s-a : s-a)$$. It suffices to prove that $$K$$ lies on the line equation of $$ID$$ where $$I$$ is the incenter. This is because $$ID \perp BC$$. Since $$I = (a:b:c)$$ we find that the line equation for $$ID$$ is $\frac{(b-c)(s-a)}{a}x + (s-b)y + (c-s)z = 0.$

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

- 2 years, 1 month ago

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