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Irish Mathematical Olympiad

Geometry Difficulty 4.5 AIME Prove it Ireland

Let $ABC$ be a triangle and $D$ , $E$ , $F$ points on the circumcircle of triangle $ABC$ so that $AD$ , $BE$ and $CF$ are diameters of the circumcircle. The lines $AE$ , $BF$ and $CD$ form a triangle $A'B'C'$ . Prove that $\triangle ABC$ is similar to $\triangle A'B'C'$ .

Solution

Let $A'$ be the intersection of $AE$ and $CD$ . Because $BE$ and $AD$ are diameters, $\angle BAA'$ and $\angle ACA'$ are right angles. Hence, $\angle BAC + \angle CAA' = 90^\circ = \angle AA'C + \angle CAA'$ . This implies $\angle BAC = \angle AA'C$ . Similarly, we obtain $\angle CBA = \angle BB'A$ and $\angle ACB = \angle CC'B$ , hence the triangles $ABC$ and $A'B'C$ are similar.

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