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.
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1$\begingroup$ Welcome to MSE. What have you tried so far? Are you stuck on some point? $\endgroup$ – Andrei Jan 15 at 10:39
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$\begingroup$ 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
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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. $\square$
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$\begingroup$ Can u help me in proving $\triangle ABC$ similar to $\triangle ARS$ ? $\endgroup$ – RB MCPE Jan 15 at 11:27
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$\begingroup$ 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
