If AB is chord of a circle with centre O, AOC is a diameter and AT is the tangent at A as shown in figure. Prove that `angleBAT=angleACB.`

Explanation

Since, AC is diameter line, so angle in semi-circle makes an angle `90^(@)`
`:.angleABC=90^(@)` [by property]
In `DeltaABC`, `angleCAB+angleABC+angleACB=180^(@)`
[:'sum of all interior angles of any triangle is `180^(@)`]
`rArrangleCAB+angleACB=180^(@)-90^(@)=90^(@)......(i)`
Since, diameter of a circle is perpendicular to the tengent.
i.e., `CAbotAT`
`:.angleCAT=90^(@)`
`rArrangleCAB+angleBAT=90^(@).....(ii)`
From Eqs. (i) and (ii).
`angleCAB+angleACB=angleCAB+angleBAT`
`rArrangleACB=angleBAT`
Hence proved.