In Fig.10.5, if AOB is a diameter of the circle and AC = BC, then CAB is equal to:
Given: AOB is the diameter of the circle.
AC = BC
⇒ ∠ABC = ∠BAC = x (say) (∵ angles opposite to equal sides are equal)
Also, diameter subtends a right angle to the circle,
∴ ∠ACB = 90°
Now, by angle sum property of a triangle, sum of all angles of a triangle is 180°.
∴ ∠CAB + ∠ABC + ∠ACB = 180°
⇒ x + x + 90° = 180°
⇒ 2x = 90°
⇒ x = 45°
∴ ∠CAB = ∠ ABC = 45°