Question 55
Let $ABC$ be a triangle with $AB = AC$ and $\angle BAC = 40^\circ$. The points $S$ and $T$ lie on the sides $AB$ and $BC$, respectively, such that $\angle BAT = \angle BCS = 10^\circ$. The straight lines $AT$ and $CS$ meet at point $P$.
Show that $BT = 2PT$.
Show that $BT = 2PT$.