MathDB

9.7

Part of 2004 All-Russian Olympiad Regional Round

Problems(1)

<MNA+<MCB=<MND+<MBC=180^o- All-Russian MO 2004 Regional (R4) 9.7

Source:

9/27/2024
Inside the parallelogram ABCDABCD, point MM is chosen, and inside the triangle AMDAMD, point NN is chosen in such a way that ∠MNA+∠MCB=∠MND+∠MBC=180o.\angle MNA + \angle MCB =\angle MND + \angle MBC = 180^o. Prove that lines MNMN and ABAB are parallel.
parallelgeometryanglesparallelogram