## Problem 2

Let $A B C D$ be a cyclic quadrilateral satisfying $A B=A D$ and $A B+B C=C D$.

Determine $\angle C D A$.


Solution 2 Answer: $\angle C D A=60^{\circ}$.

Choose the point $E$ on the segment $C D$ such that $D E=A D$. Then $C E=C D-A D=$ $C D-A B=B C$, and hence the triangle $C E B$ is isosceles.

![](https://cdn.mathpix.com/cropped/2024_06_06_070b9d2be03ba85fc7fbg-1.jpg?height=491&width=523&top_left_y=1619&top_left_x=779)

Now, since $A B=A D$ then $\angle B C A=\angle A C D$. This shows that $C A$ is the bisector of $\angle B C D=\angle B C E$. In an isosceles triangle, the bisector of the apex angle is also the perpendicular bisector of the base. Hence $A$ is on the perpendicular bisector of $B E$, and $A E=A B=A D=D E$. This shows that triangle $A E D$ is equilateral, and thus $\angle C D A=60^{\circ}$.

