Question :PA and PB are tangents drawn to the circle with centre O as shown in the figure. Prove that ZAPB = 2 ZOAB. The correct answer :As per the problem, we have the following figure: We need to prove that ZAPB = 2 ZOAB. Let us consider triangle OAB. In this triangle, we have OA = OB (radii of the same circle), and AB is a common side. Therefore, by the side-side-side (SSS) congruence criterion, we can say that triangle OAB is an isosceles triangle. Therefore, ZOAB = ZOBA. Let ZAPB = x (as given in the problem). Then, ZOAP = ZOAB - ZPAB = ZOAB - x (since PA is tangent to the circle, ZOAP = ZPAB). Similarly, ZPBO = ZOBA - ZPAB = ZOAB - x. Now, ZAOP + ZPOB + ZOAB = 180° (sum of angles of triangle AOB). Substituting the values we have obtained above, we get: (ZOAB - x) + (ZOAB - x) + ZOAB = 180° Simplifying the above equation, we get: 3ZOAB - 2x = 180° or 2x = 3ZOAB - 180° or x = (3/2)ZOAB - 90° Therefore, ZAPB = (3/2)ZOAB - 90°. Now, substituting the value of ZOAB in terms of ZAPB in the above equation, we get: ZAPB = (3/2)(ZAPB + 2ZOAB) - 90° Simplifying the above equation, we get: ZAPB = 2ZOAB Hence, proved.
