Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.9 AIME Prove it Ireland

In a triangle $ABC$ , $X$ and $Y$ are the mid-points of $AB$ and $BC$ respectively. On $BC$ there is a point $D$ , which is not the mid-point of $BC$ . Prove that $\angle XDY = \angle BAC$ implies $AD \perp BC$ .

Solution

Let $Z$ be the mid-point of $BC$ . Because $X$ , $Y$ , $Z$ are mid-points, we have $XZ \parallel AC$ , $YZ \parallel AB$ and $XY \parallel BC$ . Therefore $\angle XBZ = \angle XYZ$ and $\angle XZY$ . The latter, together with the assumption $\angle XDY = \angle BAC$ implies $\angle XDY = \angle XZY$ , hence $X$ , $D$ , $Z$ , $Y$ are concyclic.

From this we obtain $\angle XDB = \angle XYZ = \angle XBZ$ and so $|XD| = |XB| = |AX|$ . This shows that $A$ , $D$ , $B$ are on the circle with diameter $AB$ and so, by Thales' Theorem, $AD \perp BC$ .

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