try to draw it..

let AB be on the circle and segment ab be the chord..

letrbe the radius of the circle with center C..

draw the perpendicular line from C to the chord AB.. let the intersection be M.. draw triangles CAM and CBM.. note that this forms right triangles whose hypotenuses are CA and CB respectively (whose lengths are both r) .. also, they have common side CM.. you can use Pythagorean Theorem to show that AM and BM are equal in length.., that is CA^2 = CM^2 + MA^2 and CB^2 = CM^2 + MB^2, then MA = MB..