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, 2023

Geometry Difficulty 8.5 Shortlist Prove it Baltic Way

Let $\triangle ABC$ be an acute triangle with $|AB| > |AC|$ . The internal angle bisector of $\angle BAC$ intersects $BC$ at $D$ . Let $O$ be the circumcenter of $\triangle ABC$ . Let $AO$ intersect the segment $BC$ at $E$ . Let $J$ be the incenter of $\triangle AED$ . Prove that if $\angle ADO = 45^\circ$ then $|OJ| = |JD|$ .

Solution

Let $\alpha = \angle BAC$ , $\beta = \angle CBA$ , $\gamma = \angle ACB$ . We have
$\begin{align*} \angle DJA &= 90^\circ + \frac{1}{2} \angle DEA = 90^\circ + \frac{1}{2} (\angle EBA + \angle BAE) \\ &= 90^\circ + \frac{1}{2} (\beta + 90^\circ - \gamma) = 135^\circ + \frac{\beta}{2} - \frac{\gamma}{2} \end{align*}$
and
$\begin{align*} \angle DOA &= 180^\circ - \angle OAD - \angle ADO = 180^\circ - (\angle OAC - \angle DAC) - 45^\circ \\ &= 135^\circ - \left(90^\circ - \beta - \frac{\alpha}{2}\right) = 135^\circ - \left(\frac{1}{2}(\alpha + \beta + \gamma) - \beta - \frac{\alpha}{2}\right) \\ &= 135^\circ + \frac{\beta}{2} - \frac{\gamma}{2} \end{align*}$
Therefore, $\angle DJA = \angle DOA$ , hence quadrilateral $ADJO$ is cyclic.
Since $AJ$ is the bisector of $\angle OAD$ , the arcs $OJ$ and $JD$ are equal.
Hence $|OJ| = |JD|$ .

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