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BdMO Online Forum • View topic - CGMO 2012/5

CGMO 2012/5

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CGMO 2012/5

Post Number:#1  Unread postby ZM Siddiqee » Fri Mar 31, 2017 4:15 pm

Let ABC be a triangle. The incircle of ABC with center I is tangent to AB at D & AC at E. Let O denote the circumcenter of triangle BCI. Prove that, angle ODB = angle OEC.
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Re: CGMO 2012/5

Post Number:#2  Unread postby nahin munkar » Fri Mar 31, 2017 9:46 pm

$\bullet$Claim 1 : $A ,I,O$ collinear .
proof : well-known fact.

$\bullet$Claim 2:$\triangle ADO \cong \triangle AEO$
proof:

(i)$AD=AE$
(ii)$AO=AO$
(iii)$\angle DAO = \angle EAO$
$\Longrightarrow \triangle ADO \cong \triangle AEO$ $\blacksquare$

$\star$ From $Claim 2$, we get,
$\angle ODA = \angle OEA \Longrightarrow \angle ODB = \angle OEC $ $\dagger$
Last edited by nahin munkar on Fri Mar 31, 2017 10:17 pm, edited 5 times in total.
# Mathematicians stand on each other's shoulders. ~ Carl Friedrich Gauss
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Re: CGMO 2012/5

Post Number:#3  Unread postby Atonu Roy Chowdhury » Fri Mar 31, 2017 9:49 pm

It's a well known fact that $A,I,O$ are collinear. $AD=AE$ along with $\angle DAO = \angle EAO $ implies that triangle $DAO$ and $EAO$ are congruent. The rest is trivial.
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Re: CGMO 2012/5

Post Number:#4  Unread postby ahmedittihad » Fri Mar 31, 2017 10:07 pm

As both of you didn't show the proof of $A$,$I$,$O$ being colinear, I'm just showing that. Let $AI$ intersect the circumcircle of $\triangle ABC$ at $X$. We know that $X$ is the midpoint of arc $BC$. So, $BX=CX$.
Now we will show that $BX=XI$.
We have, $\angle XBI = \angle XBC + \angle CBI = \angle (A/2) + \angle (B/2) = \angle BIX$. We get that $BX=XI=XC$. So, $X$ is the circumcenter of $\triangle BIC$.
Frankly, my dear, I don't give a damn.
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