Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Provethat ∠ PTQ = 2 ∠ OPQ

Sure, here is the step by step proof:

Step 1: Draw the circle with center O. Draw the two tangents TP and TQ from the external point T to the circle.

Step 2: Join OT. Since the tangent at a point on the circle is perpendicular to the radius through the point of contact, ∠OTP = 90° and ∠OTQ = 90°.

Step 3: Join OP and OQ. Now, ∠TOP and ∠TOQ are angles in the same segment of the circle. Therefore, ∠TOP = ∠TOQ.

Step 4: In triangle OTP, ∠OTP + ∠OPQ + ∠PTQ = 180° (sum of angles in a triangle). Substituting the known values, we get 90° + ∠OPQ + ∠PTQ = 180°. Simplifying, we get ∠PTQ = 90° - ∠OPQ.

Step 5: Similarly, in triangle OTQ, ∠OTQ + ∠OQP + ∠PQT = 180°. Substituting the known values, we get 90° + ∠OQP + ∠PQT = 180°. Simplifying, we get ∠PQT = 90° - ∠OQP.

Step 6: Now, ∠PTQ = ∠PQT (from steps 4 and 5). Therefore, ∠PTQ = 2∠OPQ.

Hence, ∠PTQ = 2∠OPQ is proved.