Question

If AB is a chord of a circle with centre O, AOC is a diameter and AT is the tangent at A as shown in figure. Prove that ∠BAT = ∠ACB

Answer 1

Since, AC is a diameter line, so angle in semi-circle an angle 90°.

∴ ∠ABC = 90°   ...[By property]

In ∆ABC,

∠CAB + ∠ABC + ∠ACB = 180°  ...[∵ Sum of all interior angles of any triangle is 180°]

⇒ ∠CAB + ∠ACB = 180° – 90° = 90° ...(i)

Since, diameter of a circle is perpendicular to the tangent.

i.e., CA ⊥ AT

∴ ∠CAT = 90°

⇒ ∠CAB + ∠BAT = 90° ...(ii)

From equations (i) and (ii),

∠CAB + ∠ACB = ∠CAB + ∠BAT

⇒ ∠ACB = ∠BAT

Hence proved.