## CGMO 2012/5

- ZM Siddiqee
**Posts:**2**Joined:**Mon Mar 20, 2017 9:24 pm

### CGMO 2012/5

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.

- nahin munkar
**Posts:**55**Joined:**Mon Aug 17, 2015 6:51 pm**Location:**banasree,dhaka

### Re: CGMO 2012/5

$\bullet$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 $ $\dagger$

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

- Atonu Roy Chowdhury
**Posts:**40**Joined:**Fri Aug 05, 2016 7:57 pm**Location:**Chittagong, Bangladesh

### Re: CGMO 2012/5

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.

- ahmedittihad
**Posts:**147**Joined:**Mon Mar 28, 2016 6:21 pm

### Re: CGMO 2012/5

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 $\triangle BIC$.

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.