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equal angles starting with a parallelogram with perpenducular

Source: Mexican Mathematical Olympiad 1994 OMM P3

July 29, 2018
geometryparallelogramequal anglesperpendicular

Problem Statement

ABCDABCD is a parallelogram. Take EE on the line ABAB so that BE=BCBE = BC and BB lies between AA and EE. Let the line through CC perpendicular to BDBD and the line through EE perpendicular to ABAB meet at FF. Show that ∠DAF=∠BAF\angle DAF = \angle BAF.
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