MathDB

15

Part of 1998 Baltic Way

Problems(1)

E satisfies AE/ED=CD/DB

Source: Baltic Way 1998

1/11/2011
Given acute triangle ABCABC. Point DD is the foot of the perpendicular from AA to BCBC. Point EE lies on the segment ADAD and satisfies the equation AEED=CDDB\frac{AE}{ED}=\frac{CD}{DB} Point FF is the foot of the perpendicular from DD to BEBE. Prove that ∠AFC=90∘\angle AFC=90^{\circ}.
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