Although a smaller proof has been given here http://olympiads.hbcse.tifr.res.in/. I give another proof which is bit lengthier. In $ABC, BE$ and $CF $ intersect at $I$ the incenter. Thus $\angle BIC = 90^{\circ}+A/2$. Also $AI$ is the diameter of circle passing though $A,X,I,Y$. By sine formula in triangle $XIY$ we get $ XY= AI cos(A/2)$. Now in $AIC$ by sine formula $AI= b \frac{sin(C/2)} {cos(B/2)}$ By sine formula in $ABC$ itself gives $b+c-a = b+ b\frac{sin(C)}{sin(B)}-b\frac {sin(A)}{sin(B)}$ Just after few elementary trigonometric manipulation you end up with $(b+c-a)/2 = AI cos(A/2)$ which is same as $XY$ and we are done.

We know $BIC=90+A/2$ and we have $XY/sin(90+A/2)=AI$(I is the incenter of triangle ABC) $T$ is a point on $AB$ such that $IT$ is perpendicular to $AB$ so we have $AT=b+c-a/2=cosA/2.AI$ we know $sin(90+A/2)=cosA/2$ thus $XY=AT=b+c-a/2$ QED

My solution without trigonometry HERE, as well as many other geometry materials.

krishnamurthy95 wrote: Let $ABC$ be a triangle. Let $BE$ and $CF$ be internal angle bisectors of $\angle B$ and $\angle C$ respectively with $E$ on $AC$ and $F$ on $AB$. Suppose $X$ is a point on the segment $CF$ such that $AX$ perpendicular $CF$; and $Y$ is a point on the segment $BE$ such that $AY$ perpendicular $BE$. Prove that $XY = (b + c-a)/2$ where $BC = a, CA = b $and $AB = c$. I used ptolemy and cosine rule again and again

Mehhhh solution without trig: First notice that if $AX,AY \cap BC=J,K$, then $X$ is the midpoint of $AJ$, because it lies on the bissector and $CX\perp AJ$. Similarly for $Y$, we just need to prove $JK=2XY$, but notice $BK=BA$ and $CJ=CA$: then $JK=BC-KC-JB=a-(a-b)-(a-c)=b+c-a$, which is what we wanted.

Let $ I $ be the incenter of $ \triangle ABC $. Note that $\angle BIC =90+\angle \frac{A}{2}$ Since AXIY are concyclic, $\angle XAY = 90- \frac{A}{2} $ The circle passing through AXIY has $AI=rcosec(A/2)$ as diameter Applying sine rule in $\triangle XAY$, $\frac {XY}{sin(\angle XAY)} = 2R = AI $ $$ XY=rcosec(\frac{A}{2})cos(\frac{A}{2}) $$$$\Rightarrow XY=rcot(\frac{A}{2}) $$$$\Rightarrow \fbox {XY = s-a} $$

here is my synthetic solution

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