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Question
In the following figure, O is the centre of the circle, BD = OD and CD ⊥ AB. Find ∠CAB.
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Solution
Given, in the figure BD = OD, CD ⊥ AB.
In ΔOBD, BD = OD ...[Given]
OD = OB ...[Both are the radius of circle]
∴ OB = OD = BD
Thus, ΔODB is an equilateral triangle.
∴ ∠BOD = ∠OBD = ∠ODB = 60°
In ΔMBC and ΔMBD,
MB = MB ...[Common side]
∠CMB = ∠BMD = 90°
And CM = MD ...[In a circle, any perpendicular drawn on a chord also bisects the chord]
∴ ΔMBC ≅ ΔMBD ...[By SAS congruence rule]
∴ ∠MBC = ∠MBD ...[By CPCT]
⇒ ∠MBC = ∠OBD = 60° ...[∵ ∠OBD = 60°]
Since, AB is a diameter of the circle.
∴ ∠ACB = 90°
In ΔACB, ∠CAB + ∠CBA + ∠ACB = 180° ...[By angle sum property of a triangle]
⇒ ∠CAB + 60° + 90° = 180°
⇒ ∠CAB = 180° – (60° + 90°) = 30°
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