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: Understanding the problem:

We are given a circle with center \( O \), and two tangents \( PT \) and \( QT \) are drawn from an external point \( T \) to the circle. We are asked to prove the relationship between the angles \( \angle PTQ \) and \( \angle OPQ \).

Step 2: Using the property of tangents:

It is a well-known property that the angle between a tangent and the radius at the point of contact is always \( 90^\circ \). Since \( PT \) and \( QT \) are tangents to the circle, the following holds:
\[ \angle OTP = \angle OTQ = 90^\circ \] This means that the two angles formed between the tangents and the radii are both right angles.

Step 3: Finding the angle \( \angle PTQ \):

Next, we know that \( \angle PTQ \) is the angle between the two tangents, which is the sum of the angles \( \angle OTP \) and \( \angle OTQ \), since these angles lie along the straight line \( PTQ \). Thus, we have:
\[ \angle PTQ = \angle OTP + \angle OTQ \] Substitute the known values:
\[ \angle PTQ = 90^\circ + 90^\circ = 180^\circ \] Therefore, \( \angle PTQ \) is \( 180^\circ \).

Step 4: Establishing the relationship between the angles:

Now, let's look at the relationship between \( \angle PTQ \) and \( \angle OPQ \). We can observe that the angle \( \angle PTQ \) is twice the angle \( \angle OPQ \). This is because the angle between the tangents at the external point is twice the angle subtended by the chord \( PQ \) at the center of the circle.
Thus, we have:
\[ \angle PTQ = 2 \times \angle OPQ \] This completes the proof.

Conclusion:

We have shown that \( \angle PTQ = 2 \times \angle OPQ \), as required.