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Right triangle, feet of altitude, angle bisector and median

Source: Canada Repêchage 2015/6

June 18, 2016
geometryangle bisector

Problem Statement

Let △ABC\triangle ABC be a right-angled triangle with ∠A=90∘\angle A = 90^{\circ}, and AB<ACAB < AC. Let points D,E,FD, E, F be located on side BCBC such that ADAD is the altitude, AEAE is the internal angle bisector, and AFAF is the median.
Prove that 3AD+AF>4AE3AD + AF > 4AE.
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