Question

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

Answer 1

In the given figure,

AC is the diameter.

So, ∠CBA=90° (Angle formed by the diameter on the circle is 90º)

AT is the tangent at point A.

Thus, ∠CAT=90°

In ∆ABC,

∠BCA+∠BAC+∠CBA=180° [ Angle sum property]

⇒∠BCA+90°+∠CAT-∠BAT=180°

⇒∠BCA+90°+90-∠BAT=180°

⇒∠BCA=∠BAT

Hence Proved