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Turkish MO 1994 P2

Source: Turkish Mathematical Olympiad 2nd Round 1994

September 27, 2006
geometrycyclic quadrilateralgeometry unsolved

Problem Statement

Let ABCDABCD be a cyclic quadrilateral ∠BAD<90∘\angle{BAD}< 90^\circ and ∠BCA=∠DCA\angle BCA = \angle DCA. Point EE is taken on segment DADA such that BD=2DEBD=2DE. The line through EE parallel to CDCD intersects the diagonal ACAC at FF. Prove that AC⋅BDAB⋅FC=2. \frac{AC\cdot BD}{AB\cdot FC}=2.
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