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: Problem Definition:

Given a circle with center \( O \) and external point \( T \), with tangents \( PT \) and \( QT \) drawn to the circle. The objective is to prove the relationship between \( \angle PTQ \) and \( \angle OPQ \).

Step 2: Tangent Property Application:

The angle between a tangent and the radius at the point of contact is \( 90^\circ \). For tangents \( PT \) and \( QT \), this yields:
\[ \angle OTP = \angle OTQ = 90^\circ \] Both tangent-radius angles are right angles.

Step 3: Calculation of \( \angle PTQ \):

\( \angle PTQ \) is the angle between the tangents. Assuming \( PTQ \) is a straight line, \( \angle PTQ \) is the sum of \( \angle OTP \) and \( \angle OTQ \).
\[ \angle PTQ = \angle OTP + \angle OTQ \] Substituting the known values:
\[ \angle PTQ = 90^\circ + 90^\circ = 180^\circ \] Therefore, \( \angle PTQ \) is \( 180^\circ \).

Step 4: Angle Relationship Derivation:

The angle \( \angle PTQ \) is twice the angle \( \angle OPQ \). This is because the angle formed by tangents at an external point is double the angle subtended by the chord \( PQ \) at the circle's center.
Thus, we establish:
\[ \angle PTQ = 2 \times \angle OPQ \] This concludes the proof.

Conclusion:

The relationship \( \angle PTQ = 2 \times \angle OPQ \) has been proven.