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Slovenia — National Math Olympiad

Geometry Difficulty 4.7 AIME Prove it Slovenia

Let $\triangle ABC$ be an acute triangle with $|AB| > |AC|$ . Let $D$ be a point on the side $AB$ , such that the angles $\angle ACD$ and $\angle CBD$ are equal. Let $E$ denote the midpoint of $BD$ , and let $S$ be the circumcentre of the triangle $BCD$ . Prove that the points $A, E, S$ and $C$ lie on the same circle.

Solution

The central angle is twice the inscribed angle, so $\angle CSD = 2\angle CBD$ . The triangle $CSD$ is isosceles with the apex at $S$ , so
$\angle DCS = \frac{\pi-\angle CSD}{2} = \frac{\pi}{2} - \angle CBD.$
Thus,
$\angle ACS = \angle ACD + \angle DCS = \angle CBD + \frac{\pi}{2} - \angle CBD = \frac{\pi}{2}.$
Now, because $E$ is the midpoint of $BD$ , the line $SE$ is perpendicular to the line $BD$ and $\angle SEA = \frac{\pi}{2}$ . This implies $\angle ACS + \angle SEA = \pi$ , so the points $A, E, S$ and $C$ are concyclic.

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