Question

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

Answer 1

Given:-

$AB$ is chord.

$AOC$ is diameter

$AT$ is a tangent to the circle at $A$

To prove:-

$\angle BAT=\angle ACB$

Proof:-

$\angle OAT={90}^{\circ }$ (tangent is perpendicular to the radius )

$\Rightarrow \angle OAB+\angle BAT={90}^{\circ }-\left(1\right)$

Similarly,

$\angle CBA={90}^{\circ }$ (angle inscribed in a semicircle is ${90}^{\circ }$)

In $△ABC$

$\angle BCA+\angle CBA+\angle CAB={180}^{\circ }\Rightarrow \angle BCA+\angle CAB={90}^{\circ }-\left(2\right)$

Equating $\left(1\right)$ & $\left(2\right)$ we have

$\angle OAB+\angle BAT=\angle ACB+\angle CAB\Rightarrow \angle BAT=\angle ACB(\because \angle CAB=\angle OAB)$

Hence proved