In the given figure, O is the centre of the circle, BD = OD and CD intersect AB at 90°. Find ∠CAB.

In ∆ OBD, BD = OD

OD =OB

∴OB = OD = BD

Thus,∆ ODB is an equilateral triangle.

∴∠BOD =∠OBD =∠ODB = 60°

Now, in ∆CMB and ∆DMB, we have

MB = MB [common side]

∠CMB =∠DMB [each 90°]

CM = DM

[∵ perpendicular drawn from the centre to the chord,bisect the chord]

[by SAS congruence criterion]

Then, ∠MBC = ∠MBD [∵CPCT ]

∠MBC = ∠MBD = 60°

[∵ ∠OBD =∠MBD = 60°]

or ∠ABC = 60°

Since, AB is a diameter of the circle.

∴ ∠ACB = 90° [∵angle in a semi-circle is 90°]

In ∆ACS,∠CAB + ∠ABC + ∠ACB = 180°

[∵ sum of the angles of a triangle is 180°]

=>∠CAB + 60° + 90° = 180° [∵ ∠ABC= 60°]

∴ ∠CAB = 180°- (60° + 90°) = 30°