I have the following question:

The incircle of triangle $ABC$ touches $BC$, $CA$, $AB$ at points $D$, $E$, $F$ respectively. Suppose the incircle intersects again with $AD$ at a point $X$ such that $AX=XD$. $XB$ and $XC$ meet the incircle at points $Y$ and $Z$ respectively. Prove that $EY = FZ$.

My try:

By power of point from $A$ wrt the semicircle I have $AX\cdot AD = AE^2 = (s-a)^2$ which gives me $AX = \frac{s-a}{\sqrt{2}}$ and hence $AD = \sqrt{2}(s-a)$.

By applying Apollonius theorem in triangle $ABD$ I get $BX^2 = c(s-b)=BF\cdot BA$ which means $BX$ is tangent to circumcircle of triangle $ABX$. Similarly in triangle $AXE$.

Also by Brianchon's theorem I have that $EY$, $FZ$ and $AD$ are concurrent at a point.

However I am unsure as to how many of my obervations will be useful in solving the above problem. Can anyone help me finish the problem from here or furnish a completely new proof?

  • $\begingroup$ Won't it be like that $EY, FZ$ and $AD$ are concurrent at a point? $\endgroup$ – Anirban Niloy Mar 25 '19 at 15:19
  • $\begingroup$ yes thats right I have edited it $\endgroup$ – saisanjeev Mar 26 '19 at 4:58

We will prove that $FY\parallel EZ\parallel AD$ (it's sufficient because $EZYF$ is cyclic).

Firstly, $FY\parallel AD$ is equivalent to equality $\frac{BF}{BA}=\frac{BY}{BX}$. Since $BD$ is tangent to circumcircle of $DXY$ we have $BD^2=BX\cdot BY$. Also, $BX$ is a median in triangle $ABD$, so $$ 4BX^2=2BA^2+2BD^2-AD^2=2(AB^2+BD^2-AD\cdot AX). $$ Now note that $AF$ is tangent to circumcircle of $FXD$, so $AD\cdot AX=AF^2$. Therefore, $$ 4BX^2=2(AB^2+BD^2-AF^2)=2(AB^2+BD^2-(AB-BD)^2)=4AB\cdot BD, $$ because $BD=BF=AB-BD$ as tangents to incircle of trangle $ABC$. Hence, $BX^2=AB\cdot BD$. After this computations we obtain that $$ \frac{BY}{BX}=\frac{BX\cdot BY}{BX^2}=\frac{BD^2}{AB\cdot BD}=\frac{BD}{AB}=\frac{BF}{BA}. $$ Thus, $\frac{BF}{BA}=\frac{BY}{BX}$ and lines $FY$ and $AD$ are parallel. Similarly, $EZ$ is parallel to $AD$, so $FY\parallel EZ$. Finally, $EZYF$ is cyclic, hence $EZYF$ is an isosceles trapezoid and $EY=FZ$, as desired.

  • $\begingroup$ Sir! which software you used to make the figure? $\endgroup$ – Noor Aslam Mar 26 '19 at 5:21
  • $\begingroup$ It isn't my picture but I guess that this picture was made using Geogebra. $\endgroup$ – richrow Mar 26 '19 at 5:38
  • $\begingroup$ @richrow Hmm. Right!! $\endgroup$ – Anirban Niloy Mar 26 '19 at 6:54

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