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Junior Turkish Mathematical Olympiad

Geometry Difficulty 4.2 AIME Prove it Turkey

In a triangle $ABC$ , $|AB| = |AC|$ , $D$ is the midpoint of $[BC]$ and $E$ is the foot of the perpendicular from $D$ to the line $AC$ . Let $F$ be the second point where the line $BE$ intersects the circumcircle of the triangle $ABD$ . If $G$ is the intersection point of the lines $DE$ and $AF$ , then show that $|DG| = |GE|$ .

Solution

Let $\angle ACB = \alpha$ . Since $AB = AC$ , we have $\angle ABD = \angle ABC = \alpha$ . $DE \perp AC$ and $AD \perp BC$ imply that $\angle EDC = 90^\circ - \alpha$ and $\angle ADE = \alpha$ . Therefore, we get $\angle ABD = \angle ADE = \alpha$ which implies that $DE$ is tangent to the circuncircle of the triangle $ABD$ and hence $GD^2 = GF \cdot GA$ .

On the other hand, $\angle BFA = \angle ADB = 90^\circ$ . Therefore, $EF$ is perpendicular to the hypotenuse of the right triangle $AEG$ . Thus, $GE^2 = GF \cdot GA$ and consequently we get $GD = GE$ .

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