4. Let $A B C D$ be a convex quadrilateral in which $\angle D A C=30^{\circ}, \angle B D C=50^{\circ}, \angle C B D=15^{\circ}$ and $\angle B A C=75^{\circ}$. The diagonals of the quadrilateral intersect at point $P$. Find the measure of $\angle A P D$.

## Senior League
4. Let's construct point $E$ on diagonal $B D$ such that $\angle C E D=30^{\circ}$. Then $\angle B C E=\angle C E D-\angle C B D=15^{\circ}$. Therefore, $E B=E C$. $\angle B E C=180^{\circ}-\angle B C E-\angle C B E=150^{\circ}=2 \angle B A C$. Hence, point $A$ lies on the circle with center $E$ and radius $E C$. Therefore, $E C=E A$. Notice now that $\angle C E D=\angle C A D$, so quadrilateral $A E C D$ is cyclic. From the equality of $A E$ and $E C$, it follows that $\angle A D E=\angle E D C=50^{\circ}$. Then $\angle A P D=180^{\circ}-\angle A D E-\angle C A D=100^{\circ}$.

## Senior League