Let $ABC$ be a triangle with $AB=AC$ and $\angle BAC = 40^o$. Points $S$ and $T$ lie on the sides $AB$ and $BC$, such that $\angle BAT = \angle BCS = 10^o$. Lines $AT$ and $CS$ meet at $P$. Prove that $BT = 2PT$.
Source: 2011 Saudi Arabia Pre-TST November 1.4
Tags: equal angles, geometry, equal segments
Let $ABC$ be a triangle with $AB=AC$ and $\angle BAC = 40^o$. Points $S$ and $T$ lie on the sides $AB$ and $BC$, such that $\angle BAT = \angle BCS = 10^o$. Lines $AT$ and $CS$ meet at $P$. Prove that $BT = 2PT$.