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NMO Selection Tests for the Junior Balkan Mathematical Olympiad

Geometry Difficulty 4.5 AIME Prove it Romania

Let $ABC$ be a triangle and let $D, E, F$ be the midpoints of the sides $BC, CA, AB$ . Prove that $\angle DAC = \angle ABE$ if and only if $\angle AFC = \angle BDA$ .

Solution

Let $G$ be the centroid of the given triangle. Since $DF \parallel AC$ , one has $\angle DAC = \angle GDF$ .

If $\angle DAC = \angle ABE$ , then $\angle GDF = \angle FBG$ , hence the quadrilateral $BFGD$ is cyclic, implying $\angle AFC = \angle BDA$ . The converse holds by the same argument.

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