Question

Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that $\angle PTQ=2\angle OPQ$.

Answer 1

Prove $\angle TPQ$ and $\angle TQP$ are equal using congruence. Express $\angle TPQ$ or $\angle TQP$ in terms of $\angle OPQ$ using the fact that $\angle OPT=\angle OPQ+\angle QPT$. Then, apply the property of the sum of angles of the triangle on $\Delta TPQ$. Hence, find the relation between $\angle PTQ$ and $\angle OPQ$.

Complete step-by-step answer:

Given, PT and QT are tangents to the circle with centre O.
To prove $\angle PTQ=2\angle OPQ$.
Let us assume $\angle OPQ$ to be $x$.
We know that, OP = OQ as they are the radii of the circle.
Therefore, $\angle OPQ=\angle OQP=x\left( isosceles\Delta \right)$.
Join OT.
In $\Delta OPT$and $\Delta OQT$,
OP = OQ as they are radii of the circle.
$\angle OPT=\angle OQT=90{}^\circ $as PT and QT are tangents to the circle OT = OT (common side).
$\therefore $, by the ASS criteria of congruence,
$\Delta POT\cong \Delta QOT$.
Since $\Delta POT\cong \Delta QOT$,
$\begin{align}
  & PT=QT \\
 & \therefore \angle TPQ=\angle TQP\ \ \ \ \ \ \ \left( isosceles\ triangle \right) \\
\end{align}$
Now, let us consider $\Delta PTQ$,
$\angle PQT+\angle QPT+\angle PTQ=180{}^\circ $
But $\angle PQT=\angle QPT$(proved above)
$\begin{align}
  & \Rightarrow 2\left( \angle PQT \right)+\angle PTQ=180{}^\circ \\
 & \angle PQT=\angle OPT-\angle OPQ \\
 & \Rightarrow \angle PQT=90{}^\circ -\angle OPQ\ \ \ \ \ \ \ \ \ \ \ \left( PT\ \text{is a tangent} \right) \\
\end{align}$
Putting the value of $\angle PQT$ in the equation $2\left( \angle PQT \right)+\angle PTQ=180{}^\circ $, we get,
$\begin{align}
  & 2\left( 90{}^\circ -\angle OPQ \right)+\angle PTQ=180{}^\circ \\
 & 180{}^\circ -2\angle OPQ+\angle PTQ=180{}^\circ \\
 & \Rightarrow \angle PTQ=2\angle OPQ \\
\end{align}$
Hence proved.

Note: (1) Sum of angles of a triangle is $180{}^\circ $.
(2) Angle between the tangent and the line from the centre to the point where the tangent touches the circle is $90{}^\circ $.
(3) Angles adjacent to the equal sides of an isosceles triangle are equal.