Consider a circle with centre ‘O’ and has two parallel tangents through A & B at ends of diameter.
Let tangent through M intersect the parallel tangents at P and Q
Then, required to prove: ∠POQ = 90°.
From fig. it is clear that ABQP is a quadrilateral
∠A + ∠B = 90° + 90° = 180° [At point of contact tangent & radius are perpendicular]
∠A + ∠B + ∠P + ∠Q = 360° [Angle sum property of a quadilateral]
So,
∠P + ∠Q = 360° – 180° = 180° … (i)
At P & Q
∠APO = ∠OPQ = 1/2 ∠P ….(ii)
∠BQO = ∠PQO = 1/2 ∠Q ….. (iii)
Using (ii) and (iii) in (i) ⇒
2∠OPQ + 2∠PQO = 180°
∠OPQ + ∠PQO = 90° … (iv)
In ∆OPQ,
∠OPQ + ∠PQO + ∠POQ = 180° [Angle sum property]
90° + ∠POQ = 180° [from (iv)]
∠POQ = 180° – 90° = 90°
Hence, ∠POQ = 90°
