Question 79
Let $\triangle ABC$ be an acute triangle and let $X$ be the foot of the height drawn from $A$ and $Y$ be the intersection of the perpendicular to $AC$ drawn from $X$. If the circumcircle of triangle $ABX$ meets $BY$ at point $Z$ (distinct of $B$) and the extension of $AZ$ meets $XY$ at point $P$, then prove that $BX \cdot XP = PY \cdot XC$.