Let $ABC$ be a triangle such that $|AC|>|AB|$. Let $X$ be on line $AB$ (closer to $A$) such that $|BX|=|AC|$ and let $Y$ be on the segment $AC$ such that $|CY|=|AB|$. Intersection of lines $XY$ and bisector of $BC$ is point $P$. Prove that $\angle BPC+\angle BAC = 180^\circ$.
Wow, thanks... Now I know for almost all the problems on our TST where they came from. I'm missing the problems 2, 5, 7... do you maybe know where those came from?