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AE x FC = BF x ED for bicentric ABCD 2005 Portugal p5

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October 13, 2020
geometrytangentialcyclic quadrilateral

Problem Statement

Considers a quadrilateral [ABCD][ABCD] that has an inscribed circle and a circumscribed circle. The sides [AD][AD] and [BC][BC] are tangent to the circle inscribed at points EE and FF, respectively. Prove that AE⋅FC=BF⋅EDAE \cdot F C = BF \cdot ED. https://1.bp.blogspot.com/-6o1fFTdZ69E/X4XMo98ndAI/AAAAAAAAMno/7FXiJnWzJgcfSn-qSRoEAFyE8VgxmeBjwCLcBGAsYHQ/s0/2005%2BPortugal%2Bp5.png
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