$ABC$ is a triangle, $H$ its orthocenter, $I$ its incenter, $O$ its circumcenter and $\omega$ its circumcircle. Line $CI$ intersects circle $\omega$ at point $D$ different from $C$. Assume that $AB = ID$ and $AH = OH$. Find the angles of triangle $ABC$.
Since $AB=ID=AD=BD$, $\triangle ABD$ is an equilateral triangle. Hence $\angle ADB=60^\circ \implies \angle C=120^\circ$.
Let $R=$radius of the circumcircle. Note that $AH = 2R \cos A$, $OH^2 = 9R^2 -a^2 -b^2 -c^2$.
$AH=OH \implies 4R^2 -a^2=(2R\cos A)^2 = 9R^2 -a^2 -b^2 -c^2 \implies b^2 = 5R^2 -c^2$
Since $c=\sqrt{3}R$, we get $b = \sqrt{2}R$. Therefore, $\angle B = 45^\circ$ and so $\angle A = 15^\circ$.