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

## EXPERT ANSWER

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 = 90^{0} ……..(1)

The angle in a semi-circle is 90, so ∠ CBA = 90^{0} .

∠ CBA + ∠ OAB + ∠ACB = 18 0 ^{0} (Angle sum property of a triangle)

Therefore, 90 + ∠ OAB + x = 180^{0}

So, ∠ OAB + x = 9 0 ^{0}………….(2)

But OAB + y = 90^{0},

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

x = y.

Hence ∠ACB = ∠BAT.

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