Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.1 AIME, harder Prove it Singapore

In an acute triangle $ABC$ , $AC > AB$ , $D$ is the point on $BC$ such that $AD = AB$ . Let $\omega_1$ be the circle through $C$ tangent to $AD$ at $D$ , and $\omega_2$ the circle through $C$ tangent to $AB$ at $B$ . Let $F (\neq C)$ be the second intersection of $\omega_1$ and $\omega_2$ . Prove that $F$ lies on $AC$ .

Solution

Let $\omega_2$ intersect $AC$ at $F'$ . We shall prove $F'$ lies on $\omega_1$ . Thus $F = F'$ lies on $AC$ .

Referring to the figure on the right. First $\angle ABF' = \angle BCA = \alpha$ as $AB$ is tangent to $\omega_2$ at $B$ . Let $\angle DBF' = \beta$ . Then $\angle AF'B = \angle ABC = \angle ADB = \alpha + \beta$ so that the quadrilateral $ABDF'$ is cyclic. Thus $\angle DF'C = \angle ABC = \alpha + \beta$ .

Now refer to the figure on the left. Since $AD$ is tangent to $\omega_1$ , $\angle DFC = \angle ADB = \alpha + \beta = \angle DF'C$ . Thus $F'$ is on $\omega_1$ and we are done.

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