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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:
- BD is a diameter of the circle.
- ABC is an isosceles triangle.
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Solution
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.
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