In triangles APO and CPO: AO = OC = radius of the circle \(\angle\)APO = \(\angle\)CPO [because OP is the bisector of ∠APC] OP is common \(\therefore\) \(\triangle\)APO ≅ \(\triangle\)CPO \(\therefore\) AP = CP [corresponding sides] Similarly, by joining DB and extending OP, it can be shown that PB = DP \(\therefore\) AP + PB = CP + DP i.e., AB = CD [Proved]