If AB is a chord of a circle with centre 0, AOC is a diameter and AT is the tangent at A as shown in figure. Prove that ∠BAT = ∠ACB.
Since, AC is a diameter line, so angle in semi-circle makes an angle 90°.
$\begin{array}{lll}\therefore & \angle A B C=90^{\circ} & \text { [by property] }\end{array}$
In $\triangle A B C, \quad \angle C A B+\angle A B C+\angle A C B=180^{\circ}$
$\left[\because\right.$ sum of all interior angles of any triangle is $\left.180^{\circ}\right]$
$\Rightarrow \quad \angle C A B+\angle A C B=180^{\circ}-90^{\circ}=90^{\circ}$ $\ldots($ (i)
Since, diameter of a circle is perpendicular to the tangent.
i.e. $C A \perp A T$
$\therefore \quad \angle C A T=90^{\circ}$
$\Rightarrow \quad \angle C A B+\angle B A T=90^{\circ} \quad \ldots$ (ii)
From Eqs. (i) and (ii),
$\angle C A B+\angle A C B=\angle C A B+\angle B A T$
$\Rightarrow \quad \angle A C B=\angle B A T \quad$ Hence proved.