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<DPM = <BDC wanted, common tangent to intersecting circles related

Source: Switzerland - Swiss MO 2012 p1

July 16, 2020
equal anglesgeometrycirclestangent

Problem Statement

The circles k1k_1 and k2k_2 intersect at points DD and PP. The common tangent of the two circles on the side of DD touches k1k_1 at AA and k2k_2 at BB. The straight line ADAD intersects k2k_2 for a second time at CC. Let MM be the center of the segment BCBC. Show that ∠DPM=∠BDC \angle DPM = \angle BDC .
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