0
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Welcome to MSE. What have you tried so far? Are you stuck on some point? $\endgroup$ – Andrei Jan 15 at 10:39
  • $\begingroup$ I found out that triangle ABC should be similar to triangle ARS. But i don't know how to prove it $\endgroup$ – RB MCPE Jan 15 at 10:56
0
$\begingroup$

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. $\square$

$\endgroup$
  • $\begingroup$ Can u help me in proving $\triangle ABC$ similar to $\triangle ARS$ ? $\endgroup$ – RB MCPE Jan 15 at 11:27
  • $\begingroup$ 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. $\endgroup$ – Sota Antonino Jan 15 at 12:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.