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

(N/A) Since $AC$ is a diameter,the angle in a semi-circle is $90^{\circ}$.
Therefore,$\angle ABC = 90^{\circ}$ (Angle in a semi-circle property).
In $\Delta ABC$,the sum of interior angles is $180^{\circ}$,so $\angle CAB + \angle ABC + \angle ACB = 180^{\circ}$.
Substituting $\angle ABC = 90^{\circ}$,we get $\angle CAB + \angle ACB = 180^{\circ} - 90^{\circ} = 90^{\circ}$ .......$(i)$
Since $AT$ is a tangent at $A$ and $AC$ is the radius (part of the diameter),the tangent is perpendicular to the radius at the point of contact.
Thus,$\angle CAT = 90^{\circ}$.
This implies $\angle CAB + \angle BAT = 90^{\circ}$ ..........$(ii)$
From equations $(i)$ and $(ii)$:
$\angle CAB + \angle ACB = \angle CAB + \angle BAT$
Subtracting $\angle CAB$ from both sides,we get $\angle ACB = \angle BAT$. Hence proved.