Thers's a fairly simple computational proof (about half a page or so), but here's a proof using a bit of projective geometry: ED is the polar of C with respect to the incircle of ABC, so if CX and ED intersect at T then the points X, T, Z, C form a harmonic quadruple, and then so do the lines EX, ET, EZ, EC. These lines are intersected by line AD at X, D, Z' and A resectively, so these four points form a harmonic quadruple as well, so (XD/XA) : (Z'D/Z'A)=-1 (the segments are actually oriented segments), but since XDS/XA=-1, it means that Z'D/Z'A=1, which can only happen if Z' is the infinity point of the line AD, which means that EZ is parallel to AD. In the same way we show that FY is parallel to AD, so EZ and FY are parallel, so the quadrilateral EZYF is an isosceles trapezoid and the conclusion follows. [Moderator edit. See also http://www.mathlinks.ro/Forum/viewtopic.php?t=3496 for a stronger problem.]

By many similar triangles, ZE/EX=CE/CX=CD/CX=ZD/DX so,ZE/ZD=EX/DX=EX/AX with <EZD=<EXA, triangles EZD and EXA are similar then we have <EAX=<EDZ arcED -arcEX=arcED-arcZD arcEX=arcZD then EZ//DX so we have FY//DX//EZ then ZF=EY

Let a = BC, b = CA, c = AB be the sides and s the semiperimeter of the triangle $\triangle ABC.$ Power of the vertex A to the incircle (I) is $AX \cdot AD = AF^2 = (s - a)^2.$ Since $AX = \frac{AD}{2},$ we get $AD^2 = 2(s - a)^2.$ BX is the B-median of the triangle $\triangle ABD,$ $BX^2 = \frac{2BA^2 + 2BD^2 - AD^2}{4} = \frac{2c^2 + 2(s - b)^2 - 2(s - a)^2}{4} =$ $= \frac{c^2 + b^2 - a^2 + (a + b + c)(a - b)}{2} = \frac{c^2 + c(a - b)}{2} = c (s - b)$ Power of the vertex B to the incircle (I) is $BY \cdot BX = BD^2 = (s - b)^2.$ Thus $\frac{BY}{BX} = \frac{BY \cdot BX}{BX^2} = \frac{s - b}{c} = \frac{BF}{BA}$ It follows that $FY \parallel AD.$ In exactly the same way, we can show that $EZ \parallel AD.$ As a result, EFYZ is a cyclic trapezoid with bases $FY \parallel EZ,$ i.e., isosceles and with equal diagonals, hence $EF = YZ,\ EY = FZ.$

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what anice problem. it is easy by using harmonic. $CE^2=CD^2=CZ.CX$, then it means that $(XZED)=-1$. $BD^2=BF^2=BY.BX$, then it means that $(XYFD)=-1$. from the first we find that $E(EDZX)=-1$ and from the second we conclude that $F(XYFD)=-1$ and because $AX=XD$, $ZE$ should be parallel to $AD$. and again because $AX=XD$, $FY$ should be parallal to $AD$. and it is concluded that :$EZ,FY$ are parallel. then they forme an isoceles trapezoid in the incenter. and it is clear that the diagonals of an isoceles trapezoid are equal to each other. and it means that: $EY=FZ$ ask my later, each part u didnt figure out.

I tried to do this problem using trigonometry and I failed. We need to prove that $FY=EZ$ $\Leftrightarrow$ where $\mu = \eta$ $\mu=\angle FXY = \angle YFB$ and $\eta = \angle EXZ = \angle ZEC$. Let $\varphi = \angle DAC$, $\theta = \angle BAD$, $\xi = \angle FBY$ $\nu = \angle ECZ$ and $s=\frac{a+b+c}{2}$. We have: \[AX AD=AE AE\] \[AD = (s-a)\sqrt{2}\] From triangle ADC we get \[\frac{\sin \varphi}{\sin \gamma}= \frac{CD}{AD}\] \[\sin \varphi= \sqrt{2}\tan\frac{\alpha}{2}\cos^{2}\frac{\gamma}{2}\] Also we get \[\sin \theta = \sqrt{2}\tan\frac{\alpha}{2}\cos^{2}\frac{\beta}{2}\] In triangle AXB: \[\frac{\sin \theta}{\sin \xi}= \frac{BX}{AX}\] From triangle BFY: \[\frac{\sin \mu}{\sin \xi}= \frac{BY}{FY}\] We have: \[BX BY = BD^{2}\] \[\sin \xi = \sqrt{2}\sin \mu \sin \frac{\beta}{2}\] Also we have \[\sin \nu = \sqrt{2}\sin \eta \sin \frac{\gamma}{2}\] From Ceva' s theorem we have \[\frac{\sin \varphi}{\sin\theta}\frac{\sin \xi}{\sin \nu}=\frac{\sin(\beta-\xi)}{\sin(\gamma-\nu)}= \frac{CX}{BX}\] We need to prove that $\sin \mu = \sin \eta$ but I don't know how to finish my proof.

Let $FD\cap XY=P$ and $YF\cap BC=T$ then using Pascal's theorem on hexagon $DDXYFF$ we obtain $A, P$ and $T$ are collinear, and because $BX$ is median of $\triangle ABD$, then using Ceva's theorem we get ratios that prove that $FY\parallel AD$. Similarly on the other side we would obtain $EZ\parallel AD$, so $EZ\parallel YF$ making $YFEZ$ an iscoceles trapezium. So the diagonals are congruent hence $EY=FZ$.

It is weird this solution did not appear so far. Let $M$ be the midpoint of $FD$. We know that $XM\parallel AB$ and that $XB$ is the $X-$symmedian in $\triangle XFD$, then \[ \angle BFY=\angle FXY=\angle MXD=\angle FAD \]meaning that $FY\parallel AD$. Analogously $EZ\parallel AD$ and we are done.

here is my solution

why is an error that dont let me post latex? its says that new users cant post images