Ab and AC are two chords of a circle of radius r such that AB=2AC. If p and q are the distances of Ab and AC from the centre, prove that 4p^2= p^2 + 3r^2.

Let r be the radius of the circle. Then the segment from the center of the circle to AB, with length p, the segment from the center of AB to point B, with length AB/2, and the segment from the center of the circle to B, with length r, form a right triangle with hypotenuse of length r, legs of length p and AB/2. By the Pythagorean theorem, (AB)^2/4+ p^2= r^2. Now do the same thing with chord AC getting (AC)^2/4+ q^2= r^2. Since "AB=2AC", (AB)^2/4= (AC)^2 so we have the two equations (AC)^2+ p^2= r^2 and (AC)^2/4+ q^2= r^2.