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Geometry Difficulty 8.2 Shortlist Prove it Baltic Way

Let $\triangle ABC$ 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 $\triangle ABC$ . Prove that if $\angle IDH = \angle CBA - \angle ACB$ then $|AH| = 2 \cdot |ID|$ .

Solution

Let $H'$ be the reflection of $H$ in $BC$ . It is well-known (and easy to prove) that $H'$ lies on the circumcircle of $\triangle ABC$ . Let $O$ be the circumcenter of $\triangle ABC$ . We have
$\begin{align*} \angle OH'A &= \angle HAO = \angle BAC - \angle BAH - \angle OAC \\ &= \angle BAC - 2(90^\circ - \angle CBA) = \angle CBA - \angle ACB \\ &= \angle IDH = \angle H'HD = \angle DH'A, \end{align*}$
hence $O, D, H'$ are collinear. Also note that $\angle HAO = \angle H'HD$ implies that $AO \parallel HD$ .

Since $OM \parallel AH$ , the above equality gives that $A, O, E$ are collinear.
Let $D'$ be the reflection of $D$ in $I$ . It is well-known (and easy to prove) that $D'$ lies on $AE$ . Since $AH \parallel OM$ and $AD' \parallel HD$ , quadrilateral $AHDD'$ is a parallelogram. Therefore $|AH| = |DD'| = 2 \cdot |ID|$ .

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