Question

The following figure, shows a circle with centre ‘O’ and a tangent BPQ at point P. Show that ∠APQ + ∠BAP = 90°.

Answer 1

Given

A circle with centre O

BPQ is a tangent at point P

A and B are points on the circle

To prove:

∠APQ + ∠BAP = 90°

Step 1: Tangent–radius property

OP ⟂ BPQ (radius is perpendicular to tangent at point of contact)

∠OPQ = 90°

Step 2: Linear pair at point P

On the straight line BPQ:

∠OPQ + ∠APQ = 90°
So,
∠OPQ = 90° − ∠APQ

Step 3: Use angles in the triangle

In ΔOPA:

∠OPA + ∠OAP + ∠APA = 180°

But ∠OPA = ∠OPQ = 90° − ∠APQ

So:

(90° − ∠APQ) + ∠OAP + ∠AOP = 180°

But OA = OP (both radii), so triangle OAP is isosceles and:

∠OAP = ∠AOP

Thus:

(90° − ∠APQ) + 2∠OAP = 180°

2∠OAP = 90° + ∠APQ

∠OAP = 45° + `(∠APQ)/2`

∠BAP = ∠OAP

Substitute:

∠APQ + ∠BAP

= ∠APQ + ∠OAP

= 90°

∠APQ + ∠BAP = 90°