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The triangle $ABC$ has $CA=CB$, circumcenter $O$ and incenter $I$. The point $D$ on $BC$ is such that $DO$ is perpendicular $BI$. Show that $DI$ is parallel to $AC$.

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Is this homework? Do you have any work to show? – Henning Makholm Oct 2 '11 at 5:29
It's not a homework I have been asked by friend long back. I couldn't solve it. So I posted it here. – Ramana Venkata Oct 2 '11 at 8:22
up vote 2 down vote accepted

Hint: Since $CA=CB$, the three points $I$, $O$ and $C$ are on the same line bisecting $\angle ACB$.

Proof. Let $E$ be the intersection of $BI$ and $OD$; set $\alpha=\angle ABI$. Since $\angle EIO=90^{\circ}-\alpha$, $\angle EOI=\alpha$. $\angle IBD$ is also $\alpha$, so the four points $I$, $B$, $D$, $O$ lie on the same circle. It follows that $\angle OID=\angle OBD=\angle OCD=90^{\circ}-2\alpha$. This and $\angle ACO=90^{\circ}-2\alpha$ show that $AC$ is parallel to $DI$.

Added: The above proof is for when $O$ is between $I$ and $C$. When $I$ is between $O$ and $C$, it is necessary to change the wording a bit, but a similar argument shows that the four points $I$, $B$, $D$, $O$ lie on the same circle, from which we have $AC // DI$.

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Thanks for the answer – Ramana Venkata Oct 2 '11 at 12:00

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