Question

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

Answer 2

Answer:

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