Let $ABC$ be an isosceles triangle with $CA=CB$ and $\angle ACB=20^{\circ}$. Let $D$ be a point on side $CA$ such that $\angle ADB=30^{\circ}$. Show that $AB=CD$.
Source: Kosovo Mathematical Olympiad 2022, Grade 10, Problem 2
Tags: geometry
Let $ABC$ be an isosceles triangle with $CA=CB$ and $\angle ACB=20^{\circ}$. Let $D$ be a point on side $CA$ such that $\angle ADB=30^{\circ}$. Show that $AB=CD$.
