Given: and

In ΔOBD

OB = OD = DB

∴ Δ OBD is equilateral

∴ ∠ODB = ∠DBO = ∠BOD = 60°

Consider ΔDEB and ∠BEC

Here,

BE = BE (common)

∠CEB = ∠DEB (right angle)

CE = DE ( OE is perpendicular bisector)

∴ By SAS congruency

ΔDEB ∠BEC

∴∠DEB = ∠EBC (C.P.C.T)

∴∠EBC = 60°

Now, in ΔABC

∠EBC = 60°

∠ACB = 90° (angle in semicircle)

By angle sum property

∠EBC + ∠ACB + ∠CAB = 180°

60° + 90° + ∠CAB = 180°

∠CAB = 180° – 60° – 90° = 30°

∴∠CAB = 30°