MathDB

14

Part of 1998 Baltic Way

Problems(1)

Parallel of AC at B meets external bisector at D

Source: Baltic Way 1998

1/11/2011
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.
geometryparallelogramangle bisectorgeometry proposed