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Part of 2004 Junior Balkan Team Selection Tests - Moldova

Problems(1)

DP is angle bisector of <ADM iff PC = 4BC, ABCD parallelogram

Source: Moldova JBMO TST 2004 1.3

7/10/2020
Let ABCDABCD be a parallelogram and point MM be the midpoint of [AB][AB] so that the quadrilateral MBCDMBCD is cyclic. If NN is the point of intersection of the lines DMDM and BCBC, and P∈BCP \in BC, then prove that the ray (DP(DP is the angle bisector of ∠ADM\angle ADM if and only if PC=4BCPC = 4BC.
geometryangle bisectorCyclicparallelogram