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 formed is 90°.

i.e., ∠ABC = 90°

In ∆ABC,

∠CAB + ∠ABC + ∠BCA = 180°  .....[Angle sum property]

⇒ ∠CAB + ∠BCA = 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 equation (i) and (ii),

∠CAB + ∠ACB = ∠CAB + BAT

⇒ ∠ACB = ∠BAT