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

Unread post by Absur Khan Siam » Sun Feb 04, 2018 11:08 pm

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

Unread post by Absur Khan Siam » Sun Feb 04, 2018 11:17 pm

Applying Apollonius's theorem to $\triangle ABD$ ,
$AB^2 + AD^2 = 2AC^2 + 2BC^2 \cdots (i)$

Applying Stuart's theorem to $\triangle BCE$ ,
$CE(AB^2 + AC.CE) = BE^2.AC + BC^2.AE \cdots (ii) \Rightarrow 3AB^2 + 6AC^2 = AD^2 + 2BC^2$

$(i) + (ii)$
$\Rightarrow 4AB^2 + 4AC^2 = 4BC^2 \Rightarrow AB^2 + AC^2 = BC^2$

$\therefore \triangle ABC \text{is right-angled [Q.E.D.]}$
"(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

Unread post by thczarif » Mon Dec 03, 2018 3:26 pm

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.

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