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ABCD parallelogram, intersections

Source: Romanian Nationals RMO 2005 - grade 7, problem 1

March 31, 2005
geometryparallelogramtrigonometryAMCUSA(J)MOUSAMOangle bisector

Problem Statement

Let ABCDABCD be a parallelogram. The interior angle bisector of ∠ADC\angle ADC intersects the line BCBC in EE, and the perpendicular bisector of the side ADAD intersects the line DEDE in MM. Let F=AM∩BCF= AM \cap BC. Prove that: a) DE=AFDE=AF; b) AD⋅AB=DE⋅DMAD\cdot AB = DE\cdot DM. Daniela and Marius Lobaza, Timisoara
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