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Parallel of AC at B meets external bisector at D

Source: Baltic Way 1998

January 11, 2011
geometryparallelogramangle bisectorgeometry proposed

Problem Statement

Given triangle ABCABC with AB<ACAB<AC. The line passing through BB and parallel to ACAC meets the external angle bisector of ∠BAC\angle BAC at DD. The line passing through CC and parallel to ABAB meets this bisector at EE. Point FF lies on the side ACAC and satisfies the equality FC=ABFC=AB. Prove that DF=FEDF=FE.
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