We are considering a circle with centre ‘O’ with two parallel tangents through A & B at ends of diameter.

M intersect the parallel tangents at P and Q

Then, required to prove: $\angle POQ={{90}^{\circ }}$

Derived from diagram ABQP is a quadrilateral

$\angle A+\angle B={{90}^{\circ }}+{{90}^{\circ }}={{180}^{\circ }}$  [Tangent & radius are perpendicular at point of contact]

$\angle A+\angle B+\angle P+\angle Q={{360}^{\circ }}$  [sum property of a quadrilateral]

So,

$\angle P+\angle Q={{360}^{\circ }}-{{180}^{\circ }}={{180}^{\circ }}$  … (i)

At P & Q

$\angle OPQ=\angle APO=1/2\angle P$  …. (ii)

∠PQO = ∠BQO = 1/2 ∠Q$\angle PQO=\angle BQO=1/2\angle Q$  ….. (iii)

 From (ii) and (iii) in (i) ⇒

$2\angle OPQ+2\angle PQO={{180}^{\circ }}$

$\angle OPQ+\angle PQO={{90}^{\circ }}$  … (iv)

In $\vartriangle OPQ$ ,

$\angle OPQ+\angle PQO+\angle POQ={{180}^{\circ }}$  [sum property]

${{90}^{\circ }}+\angle POQ={{180}^{\circ }}$            [from (iv)]

$\angle POQ={{180}^{\circ }}-{{90}^{\circ }}={{90}^{\circ }}$

Hence, $\angle POQ={{90}^{\circ }}$