How to Solve this BMO 2005 round 2 problem?

Question

Let $ABC$ be a triangle with $AC > AB$. The point $X$ lies on the side $BA$ extended through $A$, and the point $Y$ lies on the side $CA$ in such a way that $BX = CA$ and $CY = BA$. The line $XY$ meets the perpendicular bisector of side $BC$ at $P$. Show that $\angle BPC + \angle BAC = 180^{\circ}$.

Answer 1

Let $l$ denote the angle bisector of $\angle{BAC}$ and $m$ the parallel to $l$ passing through the midpoint $M$ of $BC$. It suffices to prove that $XY$ is reflection of $l$ over $m$. Indeed, since midpoint of arc $BC$ of the circumcircle of $ABC$ which doesn't contain $A$ lies on $l$ and $M$ lies on $m$, the reflection of the first one with respect to the second one lies on $XY$ which happens to be a point $P$. Since the midpoint of arc satisfies the equality of angles in question and reflection with respect to $M$ preserves this property we get the result. Now, we will prove fact mentioned at the beginning it in three different ways:
1) If we denote the intersection of $l$ and $BC$ by $L$ as well as the intersection of $XY$ and $BC$ by $Z$, by the angle bisector and Thales theorem (as previously mentioned $AL || XY$) we have: $$\frac{CZ}{CY}=\frac{CL}{CA}=\frac{BL}{BA}$$ and since $CY=BA$ we have $BL=CZ$ i.e. $Z$ and $L$ are symmetric with respect to $M$ which yields the result.
2) If we consider glide reflection with respect to $m$ such that the image of $AB$ is $AC$ we notice that $X$ goes to $A$ (because $B$ goes to $C$ in this reflection as the midpoint of $BC$ stays fixed). We can consider analogous reflection taking $AC$ to $AB$. This proves that the distances of $A$, $X$ and $Y$ to $m$ are the same which finishes the proof.
3) One can also consider $X'$ and $Y'$ to be the midpoints of $AX$ and $AY$ and make use of the fact that $AX'=\frac{1}{2}AX=\frac{BX-AB}{2}=\frac{b-c}{2}$ and similarly for $AY'$ by applying Menelaus theorem to $X'$, $Y'$ and $M$ to see that they are collinear.

Answer 2

HINT.

The line through $A$ parallel to $XY$ is the bisector of angle $\angle BAC$. It meets again the circle through $ABC$ at the midpoint of arc $BC$, i.e. on the perpendicular bisector of $BC$.

Answer 3

A few things to note: $PB=PC$, since $P$ is on the perp. bisector of $BC$. Also, we have $AC=BX$ and $CY=AB$ so $AY=AX$. Also, do you see that the claim is proven if you can prove that $\angle BPC=\angle XAY$?