Prove that: BD/DC = AR/AS

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.

• Welcome to MSE. What have you tried so far? Are you stuck on some point? – Andrei Jan 15 at 10:39
• I found out that triangle ABC should be similar to triangle ARS. But i don't know how to prove it – RB MCPE Jan 15 at 10:56

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. $$\square$$
• Can u help me in proving $\triangle ABC$ similar to $\triangle ARS$ ? – RB MCPE Jan 15 at 11:27
• Ok. Firstly notice that $AD$ is perpendicular to $SR$ because quadrilateral $ASDR$ is a kite. ( $AS=SD$ and $AR=RD$ ) Now $<ASR=\frac{<ASD}{2}=<ACD=<ACB$ and for analogy $<ARS=\frac{<ARD}{2}=<ABD=<ABC$ and since we proved the equality of 2 different angles of triangles $\triangle ASR$ and $\triangle ABC$ it follows that they are similar. – Sota Antonino Jan 15 at 12:36