Prove that: BD/DC = AR/AS

Question

In a triangle ABC, internal bisector of angle A intersects side BC at D. R and S are circumcentres of triangle ABD and triangle ADC respectively. Then prove that, BD/DC = AR/AS.

Answer 1

Firstly by angle chasing we can show that the 2 isosceles triangles $\triangle ARB \equiv \triangle ASC$. Now it is well known that $\frac{BD}{DC}=\frac{AB}{AC} \ \ (1)$. But since $\triangle ARB \equiv \triangle ASC$ then $\frac{AB}{AC}=\frac{AR}{AS} \ \ (2)$. Now combining $(1)$ and $(2)$ gives us the desired result. $\square$