## EGMO 2013/1

For discussing Olympiad level Geometry Problems
Absur Khan Siam
Posts: 65
Joined: Tue Dec 08, 2015 4:25 pm
Location: Bashaboo , Dhaka

### EGMO 2013/1

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.
"(To Ptolemy I) There is no 'royal road' to geometry." - Euclid

Absur Khan Siam
Posts: 65
Joined: Tue Dec 08, 2015 4:25 pm
Location: Bashaboo , Dhaka

### Re: EGMO 2013/1

"(To Ptolemy I) There is no 'royal road' to geometry." - Euclid

thczarif
Posts: 14
Joined: Mon Sep 25, 2017 11:27 pm

### Re: EGMO 2013/1

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.