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: Understanding the Concept:
Tangents drawn from an external point to a circle are equal in length ($TP = TQ$). This creates an isosceles triangle $\triangle TPQ$. Also, the radius is perpendicular to the tangent at the point of contact ($OP \perp TP$).
Step 2: Analyzing $\triangle TPQ$:
Let $\angle PTQ = \theta$.
In $\triangle TPQ$, since $TP = TQ$, we have $\angle TPQ = \angle TQP$.
By Angle Sum Property: \[ \angle TPQ + \angle TQP + \angle PTQ = 180^\circ \] \[ 2\angle TPQ = 180^\circ - \theta \] \[ \angle TPQ = 90^\circ - \frac{1}{2}\theta \]
Step 3: Finding $\angle OPQ$:
Since $OP$ is the radius and $TP$ is the tangent, $\angle OPT = 90^\circ$. \[ \angle OPQ = \angle OPT - \angle TPQ \] \[ \angle OPQ = 90^\circ - (90^\circ - \frac{1}{2}\theta) \] \[ \angle OPQ = \frac{1}{2}\theta \]
Step 4: Final Answer:
Multiplying by 2: \[ 2\angle OPQ = \theta = \angle PTQ \] Hence proved.