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.
EGMO 2013/1
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- Posts:65
- Joined:Tue Dec 08, 2015 4:25 pm
- Location:Bashaboo , Dhaka
"(To Ptolemy I) There is no 'royal road' to geometry." - Euclid
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- Posts:65
- Joined:Tue Dec 08, 2015 4:25 pm
- Location:Bashaboo , Dhaka
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.
Again AM=AD/2=BE/2.
So, BE is the diameter of the circumcircle of triangle BAE. So,<BAE=90.So,<BAC=90.
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.