Question

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

Hint:

Remember that \( \triangle TPQ \) is isosceles and the radius is perpendicular to the tangent to link the angles.

Answer 1

Given:
From an external point T, two tangents TP and TQ are drawn to a circle with centre O.
We need to prove:
\[ \angle PTQ = 2 \angle OPQ \]

Step 1: Use properties of tangents
Tangents drawn from an external point are equal:
\[ TP = TQ \] Also, tangent ⟂ radius at the point of contact:
\[ OP \perp TP,\quad OQ \perp TQ \] So, angles at P and Q are right angles.

Step 2: Consider quadrilateral OPQT
In quadrilateral OPQT:
\[ \angle OPT = 90^\circ,\quad \angle OQT = 90^\circ \] Add these two interior angles:
\[ \angle OPT + \angle OQT = 180^\circ \] So, OPQT is a cyclic quadrilateral (opposite angles supplementary).

Step 3: Use property of a cyclic quadrilateral
In a cyclic quadrilateral, the exterior angle equals the interior opposite angle.

At point T:
Exterior angle = \(\angle PTQ\)

Interior opposite angle = \(\angle POQ\)

Therefore, \[ \angle PTQ = \angle POQ \]

Step 4: Relate central angle and angle on the circle
In the same segment of a circle:
\[ \angle POQ = 2\angle PPQ \] i.e., the angle at the centre is double the angle at the circumference standing on the same chord PQ.

So, \[ \angle POQ = 2\angle OPQ \]

Step 5: Combine the results
We already have: \[ \angle PTQ = \angle POQ \] And: \[ \angle POQ = 2\angle OPQ \] Therefore, \[ \boxed{\angle PTQ = 2\angle OPQ} \]

Hence proved.