Question

The 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: The angle between the chord and the tangent equals the angle in the alternate segment, but in this proof, focusing on the radius-tangent perpendicularity is the standard approach.

Answer 1

Step 1: Draw the required figure and identify the given elements.
Let a circle have centre O. From an external point T, two tangents TP and TQ are drawn touching the circle at points P and Q respectively. Join OP, OQ, and PQ.

Step 2: Use the property of tangents.
A radius drawn to the point of contact of a tangent is perpendicular to the tangent. Therefore:
\(OP \perp PT\) and \(OQ \perp QT\).
Thus,
\(\angle OPT = 90^\circ\) and \(\angle OQT = 90^\circ\).

Step 3: Consider quadrilateral OPQT.
The points O, P, T, Q form a quadrilateral. The sum of the interior angles of a quadrilateral is 360°.
So,
\[ \angle OPT + \angle OQT + \angle PTQ + \angle POQ = 360^\circ \]
Substitute the known right angles:
\[ 90^\circ + 90^\circ + \angle PTQ + \angle POQ = 360^\circ \]
\[ 180^\circ + \angle PTQ + \angle POQ = 360^\circ \]
\[ \angle PTQ + \angle POQ = 180^\circ \]
Therefore,
\[ \angle PTQ = 180^\circ - \angle POQ \]

Step 4: Use the property of triangle OPQ.
In triangle OPQ, since OP = OQ (radii of the same circle), triangle OPQ is isosceles.
Hence,
\[ \angle OPQ = \angle OQP \]
Using the angle sum property of a triangle:
\[ \angle OPQ + \angle OQP + \angle POQ = 180^\circ \]
\[ 2\angle OPQ + \angle POQ = 180^\circ \]
\[ \angle POQ = 180^\circ - 2\angle OPQ \]

Step 5: Substitute this value in Step 3.
\[ \angle PTQ = 180^\circ - (180^\circ - 2\angle OPQ) \]
\[ \angle PTQ = 2\angle OPQ \]

Final Result:
Hence, it is proved that
\[ \angle PTQ = 2\angle OPQ \]