In figure, O is the centre of a circle, BD = OD and CD ⊥ AB. Find ∠CAB.

Given O is the centre of the circle, BD = OD and CD ⊥ AB.

⇒ OD = OB [Radii of circle]

⇒In ΔOBD,

⇒ OD = BD = OB

∴ ΔOBD is an equilateral triangle.

Each interior angle of an equilateral triangle is 60°.

⇒∠OBD = ∠ODB = ∠BOD = 60°

Also given CD ⊥ AB,

We know that in equilateral triangle altitude drawn from any vertex bisects the vertex angle and also bisects the opposite side.

⇒∠EDB = ∠EDO = 30° … (1) and OE = BE

We know that if a perpendicular is drawn from center to any chord, it bisects the chord.

⇒ CE = DE … (2)

Consider ΔBED and ΔBEC,

⇒ CE = DE [From equation (2)]

⇒∠BED = ∠BEC = 90° [CD ⊥ AB]

⇒ BE = BE [Common side]

By SAS rule,

⇒ ΔBED ≅ ΔBEC

By CPCT,

⇒∠BDE = ∠BCE

From (1),

⇒∠BDE = ∠BCE = 30°

In ΔBCE,

From angle sum property,

⇒∠BCE + ∠BEC + ∠CBE = 180°

⇒ 30° + 90° + ∠CBE = 180°

⇒∠CBE = 180° - 120°

∴ ∠CBE = 60°

We know that the angle inscribed in a semicircle is always a right angle.

In ΔABC,

From angle sum property,

⇒∠CBA + ∠ACB + ∠CAB = 180°

⇒ 60° + 90° + ∠CAB = 180° [∠CBA = ∠CBE]

⇒∠CAB = 180° - 150°

∴ ∠CAB = 30°