
G2 Let $A B C D$ be a convex quadrilateral with $\varangle D A C=\varangle B D C=36^{\circ}, \varangle C B D=18^{\circ}$ and $\varangle B A C=72^{\circ}$. If $P$ is the point of intersection of the diagonals $A C$ and $B D$, find the measure of $\varangle A P D$.


## Solution

On the rays ( $D A$ and ( $B A$ we take points $E$ and $Z$, respectively, such that $A C=A E=$ $A Z$. Since $\varangle D E C=\frac{\varangle D A C}{2}=18^{\circ}=\varangle C B D$, the quadrilateral $D E B C$ is cyclic.

Similarly, the quadrilateral $C B Z D$ is cyclic, because $\varangle A Z C=\frac{\varangle B A C}{2}=36^{\circ}=\varangle B D C$. Therefore the pentagon $B C D Z E$ is inscribed in the circle $k(A, A C)$. It gives $A C=A D$ and $\varangle A C D=\varangle A D C=\frac{180^{\circ}-36^{\circ}}{2}=72^{\circ}$, which gives $\varangle A D P=36^{\circ}$ and $\varangle A P D=108^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_06_05_5f424aab603cc04b261bg-09.jpg?height=853&width=899&top_left_y=1047&top_left_x=584)
