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, 2023

Geometry Difficulty 8.2 Shortlist Prove it Baltic Way

Let $\triangle ABC$ be an isosceles triangle with $|AB| = |AC|$ . Let $D$ be an arbitrary point on segment $BC$ . Let $X$ and $Y$ be points on $AB$ and $AC$ , respectively, such that $XY \parallel BC$ and $XY$ passes through the midpoint of $AD$ . Prove that if the circumcenter of $\triangle ABC$ lies on $\odot(AXY)$ then the quadrilateral $AYDX$ is a parallelogram.

Solution

Note that $\angle XYO = \angle XAO = \angle OAY = \angle OXY$ , hence the triangle $\triangle OXY$ is isosceles. Let $K, L, M, N$ be the midpoints of $AB, AC, AD, XY$ respectively. Note that $K, L, M$ are collinear because they lie on the midline of $\triangle ABC$ parallel to $BC$ . Moreover, $M$ lies on $XY$ by assumption. Since $XY \parallel BC \parallel KL$ , the lines $KL$ and $XY$ do not coincide. This means that the lines $XY$ and $KL$ intersect at $M$ .

Note that $K, L, N$ are the projections of $O$ onto $AX, AY$ and $XY$ , respectively. Therefore $K, L, N$ lie on the Simson line of $O$ .

This means that $M = N$ , i.e. the segments $AD$ and $XY$ share a common midpoint. Therefore $AXDY$ is a parallelogram.

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.