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exists triangle of sides BD, DF,AF with area >1/2 ABC if<ABC =<AED, AF // BC

Source: 239 MO 2021 8-9 p5

May 2, 2021
geometrytriangle inequalitygeometric inequality

Problem Statement

The median ADAD is drawn in triangle ABCABC. Point EE is selected on segment ACAC, and on the ray DEDE there is a point FF, and ∠ABC=∠AED\angle ABC = \angle AED and AF//BCAF // BC. Prove that from segments BD,DFBD, DF and AFAF, you can make a triangle, the area of ​​which is not less half the area of ​​triangle ABCABC.
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