Given: AB is a diameter of circle with centre O. AQ, BP and PQ are tangents to the circle.
To Prove: ∠POQ = 90°
Proof:
Let the tangent PQ touch the circle at point R.
Since tangents from an external point are equal (Theorem 10.2):
From point P: BP = PR ...(1)
From point Q: AQ = QR ...(2)
Now, AB is a diameter, so AB = AQ + QB... wait — let OA = OB = radius = r, and let AB = 2r.
Consider point P: BP = PR, so ∠POQ calculation uses angles.
Using angles at O:
In △OPQ, we use the fact that:
$$\angle POB = 90° - \angle OPB \quad \text{(since OB ⊥ BP, tangent-radius)}$$
$$\angle QOA = 90° - \angle OQA \quad \text{(since OA ⊥ AQ, tangent-radius)}$$
Since ∠OBP = ∠OAQ = 90°,
$$\angle POB + \angle QOA + \angle POQ = 180° \quad \text{(angles on straight line AOB)}$$
In △OBP: $\angle POB = 90° - \angle OPB$
In △OAQ: $\angle QOA = 90° - \angle OQA$
In △OPQ, since ∠OPQ + ∠OQP + ∠POQ = 180°:
From △OBP: $\angle BOP + \angle OPB = 90°$, so $\angle BOP = 90° - \angle OPB$
From △OAQ: $\angle AOQ = 90° - \angle AQO$
Now: $\angle BOP + \angle POQ + \angle AOQ = 180°$
$$(90° - \angle OPB) + \angle POQ + (90° - \angle OQA) = 180°$$
$$\angle POQ = \angle OPB + \angle OQA$$
In △OPQ: $\angle POQ + \angle OPQ + \angle OQP = 180°$
Since $\angle OPQ = \angle OPB$ (same angle, as B, P are on same tangent) and $\angle OQP = \angle OQA$:
$$\angle POQ + \angle OPB + \angle OQA = 180°$$
But from above, $\angle POQ = \angle OPB + \angle OQA$, so:
$$\angle POQ + \angle POQ = 180°$$
$$2\angle POQ = 180°$$
$$\boxed{\angle POQ = 90°}$$
Hence proved.
Source: Chapter 10, Sections 10.2–10.3
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The key steps examiners look for:
The crucial insight is that ∠OPQ = ∠OPB and ∠OQP = ∠OQA (since PQ is a single tangent line at each external point), so the angle-sum in △OPQ combines with the straight-line condition to give 2∠POQ = 180°.