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### EGMO 2013/1

Posted: **Sun Feb 04, 2018 11:08 pm**

by **Absur Khan Siam**

The side $BC$ of $\triangle ABC$ is extended beyond $C$ to $D$ so that $CD = BC$.The side $CA$ is extended beyond $A$ to $E$ so that $AE = 2CA$.

Prove that if $AD = BE$, then $\triangle ABC$ is right-angled.

### Re: EGMO 2013/1

Posted: **Sun Feb 04, 2018 11:17 pm**

by **Absur Khan Siam**

### Re: EGMO 2013/1

Posted: **Mon Dec 03, 2018 3:26 pm**

by **thczarif**

Join D,E.Let AD meets BE at M.

Now, C is the midpoint of BD and

EA:AC=2:1. So, A if the centroid of triangle EDB . So, M is the midpoint of BE. BM=EM=BE/2.

Again AM=AD/2=BE/2.

So, BE is the diameter of the circumcircle of triangle BAE. So,<BAE=90.So,<BAC=90.