In $\Delta ABC$ , the incircle touches $AC,AB$ at the points $E,F$ respectively . $M$ is the midpoint of $EF$ . A circle $\omega$ (center $O$ ) is drawn through incenter $I$ and $M$ . Prove that $IO$ and $EF$ meet on the circle $\omega$ .

[edited. Thanks to Sowmitra .]

## A circle through incenter

### A circle through incenter

Last edited by photon on Sun Oct 26, 2014 9:04 pm, edited 1 time in total.

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### Re: A circle through incenter

Clarification please... I think there's something wrong with the question

### Re: A circle through incenter

Clearly $EF\perp AI\Rightarrow EF\perp IM$. Let $EF\cap \omega=G$. Then $\angle IMG=90^{\circ}$. So, $IG$ is a diameter of $\omega$ i.e. $IO\cap EF\in \omega$.

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