The incircle of triangle $ABC$ touches $AC$ and $BC$ respectively $P$ and $Q$. Let $N$ and $M$ be the midpoints of the sides $AC$ and $BC$ respectively.$AM$ and $BP$,$BN$ and $AQ$ intersects at the points $X$ and $Y$ respectively. If the points $C,X$ and $Y$ are collinear , then prove that $CX$ is the angle bisector of $\angle ACB$.
Problem
Source: AZE IZHO 2015 TST
Tags: geometry, incenter, angle bisector
Anar24
21.10.2016 22:32
Good question
tenplusten
25.10.2016 20:47
Here is my solution (like Istekolympiadteam said using Ceva) Let $CX\cap AB=C'$ Since $C-X-Y$ Also $CY\cap AB=C'$ Let's apply Ceva for triangle $ABC$ and cevians $CC',BN,AQ$ $\Longrightarrow$ $\frac{AC'}{BC'}=\frac{CQ}{BQ}$ Now Let's apply Ceva for triangle $ABC$ and cevians $CC',BP,AM$ $\Longrightarrow$ $\frac{AC'}{C'B}=\frac{PA}{PC}$ So $\frac{PA}{PC}=\frac{CQ}{BQ}=\frac{AC-PA}{BC-PC}$ equivalently $\frac{PA}{PC}=\frac{AC}{BC}$ and we have $\frac{AC}{BC}=\frac{PA}{PC}=\frac{AC'}{C'B}$ which means $CC'$ is the angle-bisector of $\measuredangle ACB$.
Guendabiaani
25.10.2016 21:27
Using menelaus on triangle $AQM$ we get: $$\dfrac{AY}{YQ} \cdot \dfrac{QC}{CM} \cdot \dfrac{MX}{XA} = 1.$$Using the generalized angle bisector theorem it is easy to get: $$\dfrac{AY}{YQ} = \dfrac{BC}{BQ}$$$$\dfrac{AX}{XM} = 2 \cdot \dfrac{AP}{PC}.$$By substituting the first one in the equation given by Menelaus we get $\dfrac{AX}{XM} = 2 \cdot \dfrac{QC}{QB}$. Therefore $$\dfrac{AP}{PC} = \dfrac{QC}{QB} = \dfrac{AP+QC}{PC+QB} = \dfrac{s-a+s-c}{a} = \dfrac{b}{a}.$$Therefore we have $\dfrac{AX}{XM} = \dfrac{2b}{a} = \dfrac{AC}{CM}$ which means $CX$ bisects $\angle ACM \equiv ACB$ as desired.