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Hint: Use the concept that length of tangents drawn from an external point to a circle are equal. So, TP = TQ, which are the tangents drawn from the external point T. Since, TP = TQ, so in triangle TPQ, angles TQP and TPQ will be equal due to the property that opposite angles to equal sides are equal.

__Complete step-by-step answer:__

Given in the question- A circle with centre O. Two tangents TP and TQ are drawn from external point T.

To prove: $\angle PTQ = 2\angle OPQ$.

Proof:

We know from the theorem that lengths of tangents drawn from an external point to a circle are equal. So, we can say that-

$

TP = TQ \\

\Rightarrow \angle TQP = \angle TPQ \to (1) \\

$ (Angles opposite to equal sides are equal)

Now, we know TP is tangent and OP is radius.

Therefore, $OP \bot TP$ by the theorem, tangent at any point of a circle is perpendicular to the radius through the point of contact.

So, $

\angle OPT = {90^ \circ } \\

\Rightarrow \angle OPQ + \angle TPQ = {90^ \circ } \\

\Rightarrow \angle TPQ = {90^ \circ } - \angle OPQ \to (2) \\

$

In $\vartriangle PTQ$

$\angle TPQ + \angle TQP + \angle PTQ = {180^ \circ }$ (Angle sum property of triangle)

$

\angle TPQ + \angle TPQ + \angle PTQ = {180^ \circ }(From(1):\angle TQP = \angle TPQ) \\

\Rightarrow 2\angle TPQ + \angle PTQ = {180^ \circ } \\

\Rightarrow 2({90^ \circ } - \angle OPQ) + \angle PTQ = {180^ \circ }(From(2)) \\

\Rightarrow 2({90^ \circ }) - 2\angle OPQ + \angle PTQ = {180^ \circ } \\

\Rightarrow {180^ \circ } - 2\angle OPQ + \angle PTQ = {180^ \circ } \\

\Rightarrow \angle PTQ = 2\angle OPQ \\

$

Hence Proved.

Note- Whenever such types of questions appear, then use the basic theorems of mathematics to solve the question. As mentioned in the solution, we have used the theorem that tangents drawn from an external point to a circle are equal and the theorem that the tangent at any point on a circle is perpendicular. Then in triangle PTQ, we have used the angle sum property and by using equation (1) and (2), we have proved the result.

Given in the question- A circle with centre O. Two tangents TP and TQ are drawn from external point T.

To prove: $\angle PTQ = 2\angle OPQ$.

Proof:

We know from the theorem that lengths of tangents drawn from an external point to a circle are equal. So, we can say that-

$

TP = TQ \\

\Rightarrow \angle TQP = \angle TPQ \to (1) \\

$ (Angles opposite to equal sides are equal)

Now, we know TP is tangent and OP is radius.

Therefore, $OP \bot TP$ by the theorem, tangent at any point of a circle is perpendicular to the radius through the point of contact.

So, $

\angle OPT = {90^ \circ } \\

\Rightarrow \angle OPQ + \angle TPQ = {90^ \circ } \\

\Rightarrow \angle TPQ = {90^ \circ } - \angle OPQ \to (2) \\

$

In $\vartriangle PTQ$

$\angle TPQ + \angle TQP + \angle PTQ = {180^ \circ }$ (Angle sum property of triangle)

$

\angle TPQ + \angle TPQ + \angle PTQ = {180^ \circ }(From(1):\angle TQP = \angle TPQ) \\

\Rightarrow 2\angle TPQ + \angle PTQ = {180^ \circ } \\

\Rightarrow 2({90^ \circ } - \angle OPQ) + \angle PTQ = {180^ \circ }(From(2)) \\

\Rightarrow 2({90^ \circ }) - 2\angle OPQ + \angle PTQ = {180^ \circ } \\

\Rightarrow {180^ \circ } - 2\angle OPQ + \angle PTQ = {180^ \circ } \\

\Rightarrow \angle PTQ = 2\angle OPQ \\

$

Hence Proved.

Note- Whenever such types of questions appear, then use the basic theorems of mathematics to solve the question. As mentioned in the solution, we have used the theorem that tangents drawn from an external point to a circle are equal and the theorem that the tangent at any point on a circle is perpendicular. Then in triangle PTQ, we have used the angle sum property and by using equation (1) and (2), we have proved the result.

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