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All-Soviet-Union Mathematical Olympiad

Geometry Difficulty 4.5 AIME Prove it Soviet Union

Problem:

$ABC$ is equilateral. A line parallel to $AC$ meets $AB$ at $M$ and $BC$ at $P$ . $D$ is the center of the equilateral triangle $BMP$ . $E$ is the midpoint of $AP$ . Find the angles of $DEC$ .

Solution

Solution:

Let $K$ be the midpoint of $BP$ and $L$ the midpoint of $AC$ . $EL$ is parallel to $BC$ , so $\angle ELC = 120^\circ$ . $EK$ is parallel to $AB$ , so $\angle EKC = 60^\circ$ , so $ELCK$ is cyclic. But $\angle DKC = \angle DLC = 90^\circ$ , so $DLCK$ is cyclic. Hence $D$ , $K$ , $C$ , $L$ , $E$ all lie on a circle. Hence $\angle DEC = \angle DLC = 90^\circ$ , and $\angle EDC = \angle EKC = 60^\circ$ .

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