2018 OIM Problems/Problem 6

Problem

Let $ABC$ be an acute triangle with $AC > AB > BC$. The perpendicular bisectors of $AC$ and $AB$ cut the line $BC$ at $D$ and $E$, respectively. Let $P$ and $Q$ be points other than $A$ on the lines $AC$ and $AB$, respectively, such that $AB = BP$ and $AC = CQ$, and let $K$ be the intersection of the lines $EP$ and $DQ$. Let $M$ be the midpoint of $BC$. Show that $\angle DKA = \angle EKM$.

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution

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See also

OIM Problems and Solutions