MathNet 2021 — Part 06 Question 34

In a convex quadrilateral $ABCD$, $\angle ABD = 30^{\circ}$, $\angle BCA = 75^{\circ}$, $\angle ACD = 25^{\circ}$ and $CD = CB$. Extend $CB$ to meet the circumcircle of triangle $DAC$ at $E$. Prove that $CE = BD$.