$\triangle ABC$, Let Incircle $\triangle ABC$ touch $BC,CA$ and $AB$ at $D,E,F$ respectively. Let line $AD$ cut an incircle at point $X$ and line $XB$ and $XC$ cut incircle at point $Y$ and $Z$ respectively. If $AX=XD$ ,show that $EY=FZ$
I know If $\triangle CEZ \sim \triangle CAX$ and $\triangle BFY \sim \triangle BAX$ , then $EZ//AD//FY$. It follow that $EY=FZ$ . But I have no idea to prove that $\triangle CEZ \sim \triangle CAX$ and $\triangle BFY \sim \triangle BAX$.
This is actually a problem B2 from 10th Iberoamerican Mathematical Olympiad (1995). Several different solutions can be found at AOPS:
https://artofproblemsolving.com/community/c6h4954p15699
Also don't miss the solution given on a different page (it could well be the simplest):
https://artofproblemsolving.com/community/c6h3496