Question

Tangent AP and AQ are drawn to circle with centre O, from an external point A. Prove that $\angle PAQ=2.\angle OPQ$.

Answer 1

Consider the above drawn figure:

$OP=OQ$ (Radii of same circle)

Thus, $\angle OPQ=\angle OQP$ (Angle opposite to equal sides are equal)

In $△OPQ$, the sum of the angles is ${180}^0$ that is:

$\angle OPQ+\angle OQP+\angle POQ={180}^0$

But $\angle OPQ=\angle OQP$, therefore,

$\angle OPQ+\angle OQP+\angle POQ={180}^0\Rightarrow \angle OPQ+\angle OPQ+\angle POQ={180}^0\Rightarrow 2\angle OPQ+\angle POQ={180}^0\Rightarrow 2\angle OPQ={180}^0-\angle POQ........\left(1\right)$

We also know that

$\angle POQ+\angle PAQ={180}^0\Rightarrow \angle PAQ={180}^0-\angle POQ........\left(2\right)$

From equations 1 and 2, we get

$2\angle OPQ=\angle PAQ$

Hence, $\angle PAQ=2\angle OPQ$.