+++++++++ +++ +++ +S ++ + \ + +  \ + +  \ + +  \ _Q +  \ O _ /+ +  \ _ / + +  \ _ / + AXCYB +  _ \ / + ++ _ \ / ++ P+++ \ /++ ++++++++R 
In circle O, a chord AB is drawn, with midpoint C. Through C, two more
chords are drawn, PQ and RS. Connecting the endpoints, PS passes through
AC at point X, and QR passes through BC at point Y. Show that CX = CY.
First try showing:

Source: The rec.puzzles newsgroup long ago, though I can't find it in the rec.puzzles archive.
From Philippe Fondanaiche, Paris, France I have lost the recollection of the basical theorems of geometry.I have found a solution of your problem, I admit, not very elegant: 1) RS is a diameter of the circle Let OM=d and OR=r The law of sines in the triangle MPR gives: sin(MPR)/MR = sin(PMR)/PR therefore sin(MPR)/(d+r) = sin(MPR+MRP)/(2rcos(MRP)) hence (r+d) tg(MRP) = (rd) tg(MPR) as MPR = MSY then MX =MY 2) any R on the circle Let AMB the xaxis and MO the yaxis Let OR = r = 1(to simplify) Let OM = d <1 Equations of line PQ y=ax a>0 of line RS y=bx b<0 of circle O x^2 + (yd)^2 = 1 Let A = sqroot(1d^2+a^2) and B=sqroot(1d^2+b^2) The coordinates of the vertices P,Q,R,S are the following: P x_p = (da+A)/(1+a^2) y_p = ax_p Q x_q = (daA)/(1+a^2) y_q = ax_q R x_r = (dbB)/(1+b^2) y_r = bx_r S x_s = (db+B)/(1+b^2) y_s = bx_s MX=  MY is equivalent to x_p.x_s/(y_p  y_s) =  x_q.x_r/(y_q  y_r) So we hane to demonstrate the following relation (R): a.x_p.x_q.(x_r+x_s) = b.x_r.x_s.(x_p+x_q) Whereas x_p.x_q = (1d^2)/(1+a^2) x_p + x_q = 2da/(1+a^2) and x_r.x_s = (1d^2)/(1+ b^2) x_r + x_s = 2db/(1+b^2) the relation (R) is easily proved.