NMO Selection Tests

Geometry Difficulty 7.6 National Olympiad, round 2 Prove it Romania

Let $\triangle ABC$ be an isosceles triangle with $\angle BAC > 90^\circ$ , and let $C$ be the circle centered at $A$ with radius $AB$ . Let $M$ be the midpoint of side $AC$ . The line $BM$ intersects the circle $C$ a second time at point $D$ . Let $E$ be a point on the circle $C$ such that $BE \perp AC$ , and suppose that $DE \cap AC = \{N\}$ . Show that:
$AN = 2 \cdot AB.$

Solution

From the condition $AC \perp BE$ , it follows that $AC$ is the perpendicular bisector of segment $BE$ , so $MB = ME$ . Since $AB = AE$ , we conclude that $\triangle MAB \equiv \triangle MAE$ (congruent triangles). It follows that $\angle MAB = \angle MAE$ (1), and $\angle MBA = \angle MEA$ (2).
From $AB = AD$ , it follows that triangle $ABD$ is isosceles with base $BD$ , so $\angle ABD = \angle ADB$ (3).
From (2) and (3), we deduce that $\angle MAE = \angle MDA$ , so quadrilateral $MAED$ is cyclic. It follows that $\angle AED = \angle DMN$ , and since $\angle AMB = \angle DMN$ , we obtain $\angle AED = \angle AMB$ (4).
From (1) and (4), it follows that $\triangle MAB \sim \triangle AEN$ , so $\frac{AN}{AB} = \frac{AE}{AM} = 2$ , from which we conclude $AN = 2 \cdot AB$ .

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