Question

In the given figure, PQ is a tangent to the circle at A. AB and AD are bisectors of ∠CAQ and ∠PAC. if ∠BAQ = 30°. Prove that:

1. BD is a diameter of the circle.
2. ABC is an isosceles triangle.

Answer 1

i) ∠BAQ = 30°

Since AB is the bisector of ∠CAQ

= ∠CAB = ∠BAQ = 30°

⇒ ∠CAQ = 2 × 30° = 60°

Since P–A–Q is a straight line, ∠CAP + ∠CAQ = 180°

⇒ ∠CAP + 60° = 180°
⇒ ∠CAP = 120°

AD is the bisector of ∠CAP

⇒ ∠CAD = `1/2` × 120° = 60°

So ∠CAD + ∠CAB = 60° + 30° = 90°

Since the angle in a semi-circle = 90°

= Angle made by the diameter to any point on the circle is 90°

So, BD is the diameter of the circle.

Since BD is the diameter of the circle, it will pass through the centre.

By the Alternate segment theorem

∠ABD = angle DAC = 60°

So, in ∠BMA,

∠AMB = 90° (because MA is perpendicular to diameter BD)

= angle BMA = angle BMC = 90°

In ΔBMA and ΔBMC:

∠BMA = ∠BMC = 90°

BM = BM (common side)

MA = MC (both are radii of the circle)

So, ΔBMA ≅ ΔBMC

⇒ AB = BC (SAS congruence criterion)

∴ ΔABC is an isosceles triangle.