Question

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 $AX\perp CF$, and $Y$ is a point on the segment $BE$ such that $AY\perp BE$.

Prove that $XY=\frac{(b+c-a)}{2}$, where $BC=a$, $CA=b$, and $AB=c$.

Answer 1

Produce AX and AY to meet BC is X' and Y' respectively.
Since BY bisects $\angle AB{Y}^{\prime }$ and $BY\perp A{Y}^{\prime }$ it follows that BA = BY' and AY = YY'.
Similarly, $CA=C{X}^{\prime }$ and $AX=X{X}^{\prime }$.
Thus X and Y are mid-points of AX' and AY' respectively. By mid-point theorem XY = X'Y'/2. But $X^{\prime }{Y}^{\prime }=X^{\prime }C+Y^{\prime }B-BC=AC+AB-BC=b+c-a$
Hence $XY=(b+c-a)/2$.