a_507_bc wrote: Let $ABC$ be an acute triangle with $AB<AC$ and incenter $I$. Let $D$ be the projection of $I$ onto $BC$. Let $H$ be the orthocenter of $ABC$ and suppose that $\angle IDH=\angle CBA-\angle ACB$. Prove that $AH=2ID$. Let $P$ be a point on the incircle of $ABC$ such that $DP\parallel AI$. Then by angle chasing we have $\angle DPI=\angle IDP=\frac{\beta-\gamma}{2}$. So $DP$ is the internal angle bisector of $\angle IDH$. Let $T$ be the reflexion of $H$ w.r.t. $BC$. Then $T$ is on the circumcircle of $ABC$ and we have $\angle DTA=\beta-\gamma$. So $TD$ passes through the reflection $X$ of $A$ w.r.t. the perpendicular bisector of $BC$. But $XD$ passes through the tangency point of the circumcircle and the $A$-mixtilinear incircle. So $T$ in this tangency point and $P$ lies on $AT$. So $P$ lies on $AH$. Since $AH\parallel ID$ have have $\angle HPD=\angle IDP$. So $DIPH$ is a rhombus. But now we have $HI\perp DP\parallel AI$. So $\angle AIH=90^\circ$ and the midpoint of $AH$ is the circumcenter of $AHI$. But $DP$ is the perpendicular bisector of $HI$. Thus $P$ is the circumcenter of $AHI$ and the midpoint of $AH$. Thus $AH=2PH=2ID$.

This problem was proposed by Burii.

Let $H'$ be the reflection of $H$ across $BC$. It is known that $H'$ lies on the circumcircle of $\triangle ABC$. $$\angle OAC = 90 - \frac{1}{2}\angle AOC = 90 - \angle CBA$$$$\angle OH'A = \angle H'AO$$$$= \angle BAC - (\angle BAH + \angle OAC)$$$$=\angle BAC - 2(90-\angle CBA)$$$$=2\angle CBA + (180-\angle CBA- \angle BCA) - 180$$$$= \angle CBA - \angle BCA$$$$= \angle IDH = \angle DHH' = \angle DH'H$$Hence $O,D,H'$ are collinear, $OH//DH$. Let the point of intersection of extended $DI$ and $OA$ be $D'$. Let $O'$ the projection of $O$ onto $BC$. $$\angle O'OE = \angle OAH = \angle OH'A = \angle O'OD $$$OD = OE$, i.e. $O'$ is the midpoint of $DE$. $$\angle OD'D =\angle OAH =\angle DHH' =\angle IDH =\angle DIO' $$$$\rightarrow IO'//DE$$ Hence, $I$ is the midpoint of $DD'$. $AHDD'$ is a parallelogram. $AH = 2ID$.