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Geometry Difficulty 4.9 AIME Prove it Argentina

Let ABCDABCD be a parallelogram whose diagonals meet at MM. Let NN be an interior point of triangle AMBAMB such that ∠AND=∠BNC\angle AND = \angle BNC. Prove that ∠MNC=∠NDA\angle MNC = \angle NDA and ∠MND=∠NCB\angle MND = \angle NCB.

Solution

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

Figure 1

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