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Problem 2313

IMO Shortlist mid-range; USAMO P2/P5

Geometry Difficulty 8.5 Prove it Romania — NMO Selection Tests for the Balkan and International Mathematical Olympiads · Romania

Let $ABC$ be a triangle such that $AB < AC$ . The perpendicular bisector of the side $BC$ meets the side $AC$ at the point $D$ , and the (interior) bisectrix of the angle $ADB$ meets the circumcircle $ABC$ at the point $E$ . Prove that the (interior) bisectrix of the angle $AEB$ and the line through the incentres of the triangles $ADE$ and $BDE$ are perpendicular.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

The lines $BC$ and $DE$ are parallel, so the angles $BED$ and $DAE$ are equal. Then so are the angles $AED$ and $DBE$ . Let $I$ and $J$ be the incentres of the triangles $ADE$ and $BDE$ , respectively. It follows that the triangles $DIE$ and $DJB$ are similar, so $DI/DE = DJ/DB$ . Since the angles $IDJ$ and $EDB$ are equal, the triangles $DIJ$ and $DEB$ are similar, so the angles $DIJ$ and $DEB$ are equal. Let the line $BE$ meet the line $IJ$ at the point $F$ . Notice that the quadrangle $DIEF$ is cyclic to deduce that the angles $EFI$ and $EDI$ are both equal to one half of the angle $ACB$ . Consequently, the line $IJ$ is parallel to the exterior bisectrix of the angle $AEB$ . The conclusion follows.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.