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 : BD is diameter of the circle.

From figure,

∠CAB = ∠BAQ = 30° (AB is angle bisector of ∠CAQ)

⇒ ∠CAQ = 2∠BAQ = 60°.

From figure,

⇒ ∠CAQ + ∠PAC = 180° [Linear pair]

⇒ 60° + ∠PAC = 180°

⇒ ∠PAC = 180° - 60°

⇒ ∠PAC = 120°.

⇒ ∠PAC = 2∠CAD (AD is angle bisector of ∠PAC)

⇒ 120° = 2∠CAD

⇒ ∠CAD = $\dfrac{120°}{2}$

⇒ ∠CAD = 60°.

From figure,

∠DAB = ∠CAD + ∠CAB = 60° + 30° = 90°.

Thus BD, subtends 90° on the circle. Since, angle in semi-circle is a right angle.

Hence, BD is the diameter of the circle.