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Perpendicular projections and acute-angled triangle

Source: Poland 2004

October 14, 2004
trigonometrygeometryvectorinequalitiescalculusderivativegeometry solved

Problem Statement

3. In acute-angled triangle ABC point D is the perpendicular projection of C on the side AB. Point E is the perpendicular projection of D on the side BC. Point F lies on the side DE and: EFFD=ADDB\frac{EF}{FD}=\frac{AD}{DB} Prove that CF⊥AECF \bot AE
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