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BD/CD<HD/MD in right triangle, altitude, bisector, median from A

Source: 1990 Greece MO Grade XI p1

September 6, 2024
geometryangle bisectorgeometric inequalityright triangle

Problem Statement

Let ABCABC be a right triangle with ∠A=90o\angle A=90^o and AB<ACAB<AC. Let AH,AD,AMAH,AD,AM be altitude, angle bisector and median respectively. Prove that BDCD<HDMD.\frac{BD}{CD}<\frac{HD}{MD}.
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