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Turkey TST 1992 - P4: PQ+RS=QR+SP in cyclic quadrilateral

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March 13, 2011
geometrycyclic quadrilateralgeometry proposed

Problem Statement

The feet of perpendiculars from the intersection point of the diagonals of cyclic quadrilateral ABCDABCD to the sides AB,BC,CD,DAAB,BC,CD,DA are P,Q,R,SP,Q,R,S, respectively. Prove PQ+RS=QR+SPPQ+RS=QR+SP.
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