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Topic: part one of the hard question trigiolgy
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Posts: 16
Registered: 4/30/07
part one of the hard question trigiolgy
Posted: Jun 15, 2007 2:19 AM
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P, Q are points on the side AC, AB of triangle ABC such that
angBPC = ang BQC; BP cuts CQ at K; X, Y are points on AC, AB such that KXAY is a parallelogram. Prove that (i) AX.XC = AY.YB (ii) the centre of the circle ABC is equidistant from X and Y

I hav already solve (i) but just to typed if it is of any help to (ii)
PS sorry but i dont know how to draw diagrams on computer

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