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IMO 3J, Independent Study 2

Geometry Difficulty 8.7 Shortlist Prove it Taiwan

Let $ABC$ be a triangle such that the angular bisector of $\angle BAC$ , the $B$ -median and the perpendicular bisector of $AB$ intersect at a single point $X$ . Let $H$ be the orthocenter of $ABC$ . Show that $\angle BXH = 90^\circ$ .

Solution

Let $M$ be the midpoint of $AC$ , and $X'$ be the reflection of $X$ with respect to $M$ . Since $XA = XB$ and $AX$ is the angular bisector of $\angle BAC$ , we know that the circumcircle of $AXB$ is tangent to $AC$ and so $MA^2 = MX \times MB$ as $M = BX \cap AC$ . Therefore $MA \times MC = MX \times MB$ and so $X'$ lies on the circumcircle of $ABC$ . Note that the reflection $H'$ of $H$ with respect to $M$ is the antipodal point of $B$ , showing that $HX \parallel H'X' \perp BX \parallel BX$ , as desired.

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from zh; metadata (topic, difficulty) added by this project.