Now, we can easily get that $\bigtriangleup ABD \cong \bigtriangleup ACE$ and $\bigtriangleup BCD \cong \bigtriangleup CDE$
this gives us that $ABCP$ and $APDE$ is concyclic. So, $\angle CBD = \angle CAM = \alpha$ and $\angle DAM = \gamma$. Now.sine law on $\bigtriangleup CDY$ and $\bigtriangleup CYA$ gives us
Now, let $AP \cap CD = M$. Ceva's theorem on $\bigtriangleup ACD$ gives us $\dfrac{CM}{MD} = 1$, or,
$CM = MD$
Re: ISL 2006 G3
Posted: Sat May 20, 2017 12:45 am
by Atonu Roy Chowdhury
Solution with spiral similarity is easy. Here I'm gonna give a different solu. And I chased some angles and lengths.
$\triangle AED$ and $\triangle ABC$ are similar and so $AC.AD=AB.AE$ which along with $\angle BAD = \angle CAE$ implies $\triangle ABD$ and $\triangle ACE$ are similar and so $ABCP$,$AEDP$ are cyclic. Again trivial angle chasing yields $\angle CAM = \angle PCM$ where $M$ is the intersection point of $AP$ and $CD$ . So, $\triangle CAM$ and $\triangle PCM$ are similar. That implies $CM^2=AM.PM$. Similarly $DM^2=AM.PM$ .
QED