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jason
Posts:
16
Registered:
4/30/07


part one of the hard question trigiolgy
Posted:
Jun 15, 2007 2:19 AM


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



