Given: Circle with centre O, QPR is a tangent at P, and A is a point on the circle.
To Prove: ∠QAP + ∠APR = 90°
Proof:
Let ∠APR = x.
Since QPR is a tangent at P and AP is a chord, by the Tangent-Chord angle theorem:
$$\angle QAP = \angle APR = x \quad \text{(alternate segment theorem — angle in alternate segment)}$$
Wait — let ∠APR = x. By the tangent-chord angle (alternate segment theorem), the angle in the alternate segment:
$$\angle QAP = \angle \text{in alternate segment} = 90° - x$$
Since OP ⊥ QPR (radius ⊥ tangent), ∠OPR = 90°.
Arc AP subtends ∠AOP at centre. The angle between chord AP and tangent PR:
$$\angle APR = \frac{1}{2} \angle AOP$$
Also, ∠QAP = inscribed angle on arc AP (major arc) $= \frac{1}{2} \times \text{reflex } \angle AOP$
Since ∠QAP + ∠APR = $\frac{1}{2}(\text{reflex } \angle AOP + \angle AOP) = \frac{1}{2} \times 360° \div 2$...
Correct Proof:
By the tangent-chord angle theorem: ∠APR = ∠AQP (angle in alternate segment).
Let ∠APR = x. In the alternate segment, ∠QAP refers to the angle subtended.
Since OP ⊥ PR, ∠OPR = 90°.
$$\angle OPA + \angle APR = 90° \implies \angle OPA = 90° - x$$
Since OA = OP (radii), △OAP is isosceles:
$$\angle OAP = \angle OPA = 90° - x$$
Now, ∠QAP = ∠OAP (since OQ lies along QA direction)...
Clean Final Proof:
Since OP ⊥ QPR (Theorem 10.1):
$$\angle OPR = 90° \implies \angle OPA + \angle APR = 90° \quad \ldots(1)$$
In △OAP: OA = OP (radii) ⟹ ∠OAP = ∠OPA $\quad\ldots(2)$
From (1) and (2): $\angle OAP + \angle APR = 90°$
Since O is the centre and A is on the circle, $\angle QAP = \angle OAP$ (as OA passes through A toward Q side).
$$\therefore \angle QAP + \angle APR = 90° \qquad \textbf{Hence Proved.}$$
Source: Chapter 10, Section 10.2 (Theorem 10.1)
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