Question 60
Let $P$ be the intersection of the diagonals of a convex quadrilateral $ABCD$. Let $X,Y,Z$ lie in the interiors of $AB,BC,CD$, respectively, with
$$\frac{|AX|}{|XB|}=\frac{|BY|}{|YC|}=\frac{|CZ|}{|ZD|}=2.$$
Suppose also that $XY$ is tangent to the circumcircle of $\triangle CYZ$ and $YZ$ is tangent to the circumcircle of $\triangle BXY$. Prove that $\angle APD=\angle XYZ$.
$$\frac{|AX|}{|XB|}=\frac{|BY|}{|YC|}=\frac{|CZ|}{|ZD|}=2.$$
Suppose also that $XY$ is tangent to the circumcircle of $\triangle CYZ$ and $YZ$ is tangent to the circumcircle of $\triangle BXY$. Prove that $\angle APD=\angle XYZ$.