MathDB
AQ+CQ=BP - Iran NMO 1998 (Second Round) Problem5

Source:

October 4, 2010
geometrycircumcircleperpendicular bisectorcyclic quadrilateralgeometry proposed

Problem Statement

Let ABCABC be a triangle and AB<AC<BCAB<AC<BC. Let D,ED,E be points on the side BCBC and the line ABAB, respectively (AA is between B,EB,E) such that BD=BE=ACBD=BE=AC. The circumcircle of ΔBED\Delta BED meets the side ACAC at PP and BPBP meets the circumcircle of ΔABC\Delta ABC at QQ. Prove that: AQ+CQ=BP. AQ+CQ=BP.
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