Question

If $AB$ is a chord of a circle with centre ${}^{\prime }{O}^{\prime }AOC$ is a diameter and $AT$ is the tangent is the tangent at A as shown in Figure. Prove that $\angle BAT=\angle ACB.$

Answer 1

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