Two circles intersect at A and B. From a point P on one of the circles, two line segments PAC and PBD are drawn intersecting the other circle at C and D respectively. then CD is.

Draw MP is tangent at P Joint AB

$\angle MPA=\angle PBA$ (Angles in the alternate Segment)

$\angle PBA=\angle ACD$ (Exterior angle of a Cyclic Quadrilateral ABCD is equal to its interior opposite angle)

$\therefore \angle MPA=\angle ACD$

$\Rightarrow \angle MPC=\angle PCD$

But these are a pair of alternative interior angles

$\therefore$ MP||CD