If AB is a chord of a circle with centre O, AOC is diameter and AT is the tangent at the point A.

Prove that ∠BAT=∠ACB

Let ∠ACB = x and ∠BAT = y.

A tangent makes an angle of 90 degrees with the radius of a circle,

so we know that ∠OAB + y = 90⁰……..(1)
The angle in a semi-circle is 90, so ∠CBA = 90⁰.
∠CBA + ∠OAB + ∠ACB = 180⁰  (Angle sum property of a triangle)
Therefore, 90 + ∠OAB + x = 180⁰

So, ∠OAB + x = 90⁰………….(2)
But OAB + y = 90⁰, 

Therefore, ∠OAB + y = ∠OAB + x  ………….[From (1) and (2)]
x = y.

Hence ∠ACB = ∠BAT.