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Rioplatense Mathematical Olympiad

Geometry Difficulty 4.9 AIME Prove it Argentina

Let $ABCD$ be a parallelogram whose diagonals meet at $M$ . Let $N$ be an interior point of triangle $AMB$ such that $\angle AND = \angle BNC$ . Prove that $\angle MNC = \angle NDA$ and $\angle MND = \angle NCB$ .

Solution

Let $N'$ be the symmetric of $N$ relative to $M$ . Since the diagonals cut in half, $AN'CN$ and $DN'BN$ are parallelograms. Hence, $\angle DN'A = \angle BNC$ and $\angle BN'C = \angle DNA$ . As $\angle DNA = \angle BNC$ , then the quadrilaterals $DN'NA$ and $BNN'C$ are cyclic. Thus, $\angle MNC = \angle N'NC = \angle AN'N$ ( $AN' \parallel NC$ ) and $\angle AN'N = \angle NDA$ (cyclic $DN'NA$ ), hence $\angle MNC = \angle NDA$ . Similarly, $\angle MND = \angle NCB$ .

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