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Geometry Difficulty 7.3 National Olympiad, round 2 Prove it Hong Kong

Given a triangle $ABC$ , let $M$ be the midpoint of $BC$ . The circle passing through $A$ and tangent to $BC$ at $M$ cuts $AB$ and $AC$ at $D$ and $E$ respectively. Suppose $B, C, E$ and $D$ are concyclic. Show that $AB = AC$ .

Solution

Firstly, consider the power of $B$ with respect to $(ADME)$ . This gives
$BD \times BA = BM^2.$
Similarly, we get
$CE \times CA = CM^2$
by considering the power of $C$ with respect to the same circle. As $M$ is the midpoint of $BC$ , the two expressions are equal, so that
$BD \times BA = CE \times CA. \qquad (1)$
Next, since $B, C, E, D$ are concyclic, we have
$AD \times AB = AE \times AC \qquad (2)$
by considering the power of $A$ . Adding (1) and (2), we get
$AB^2 = AB \times BD + AB \times AD = AC \times CE + AC \times AE = AC^2.$
This gives $AB = AC$ .

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