In $\Delta ABC$,the bisectors of $\angle B$ and $\angle C$ intersect at $P$. $A$ line drawn through $P$ and parallel to $BC$ intersects $AB$ at $X$ and $AC$ at $Y$. Prove that $XY = XB + YC$.

(N/A) Given that $BP$ is the bisector of $\angle B$,therefore $\angle XBP = \angle PBC$.
Since $XY \parallel BC$,the alternate interior angles are equal,so $\angle PBC = \angle BXP$.
Thus,$\angle XBP = \angle BXP$.
In $\Delta XBP$,since the base angles are equal,the opposite sides are equal,so $XB = XP$.
Similarly,$CP$ is the bisector of $\angle C$,so $\angle YCP = \angle PCB$.
Since $XY \parallel BC$,the alternate interior angles are equal,so $\angle PCB = \angle CYP$.
Thus,$\angle YCP = \angle CYP$.
In $\Delta YCP$,since the base angles are equal,the opposite sides are equal,so $YC = YP$.
Now,$XY = XP + YP$.
Substituting the values,we get $XY = XB + YC$.
Hence proved.