# Help in solving this geometry problem

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?

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

## 1 Answer

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.

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