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