Question

In the given figure, two tangents PT and QT are drawn to a circle with centre O from an external point T. Prove that \(\angle\) PQT = \(2\angle\) OPQ.

Answer 1

Step 1: Introduction to the problem:
We are given a circle with center $O$, and two tangents $PT$ and $QT$ drawn from the external point $T$ to the circle.
Since $PT$ and $QT$ are tangents, the angle between each tangent and the radius of the circle is always $90^\circ$.

Step 2: Understanding the property of tangents and the radius:
- The radius $OT$ is perpendicular to the tangent $PT$ at the point of contact. Hence, we have:
\[ \angle OTP = 90^\circ \] - Similarly, the radius $OT$ is perpendicular to the tangent $QT$ at the point of contact. Hence, we have:
\[ \angle OTQ = 90^\circ \]

Step 3: Finding the angle $\angle PTQ$:
We are given that the angle $\angle PTQ$ is the angle formed between the two tangents. We know that this angle is the sum of $\angle OTP$ and $\angle OTQ$, since they are adjacent angles formed by the tangents and the radii.
Thus, we have:
\[ \angle PTQ = \angle OTP + \angle OTQ \] Substitute the values for $\angle OTP$ and $\angle OTQ$:
\[ \angle PTQ = 90^\circ + 90^\circ = 180^\circ \]

Step 4: Conclusion - Relating $\angle PTQ$ and $\angle OPQ$:
Since $OT$ is the radius and $PT$ and $QT$ are tangents, $\angle PTQ$ is formed by the two tangents, and it is equal to twice the angle $\angle OPQ$ (as $OPQ$ is half the angle between the two tangents). Therefore, we conclude that:
\[ \angle PTQ = 2 \angle OPQ \] Thus, we have shown that the angle between the tangents is twice the angle subtended by the chord at the center of the circle.