$ A B $ is a chord of a circle with centre $ O, A O C $ is a diameter and $ A T $ is the tangent at $ A $ as shown in the figure. Prove that $ \angle B A T=\angle A C B $."

Given:

\( A B \) is a chord of a circle with centre \( O, A O C \) is a diameter and \( A T \) is the tangent at \( A \) as shown in the figure.

To do:
We have to prove that \( \angle B A T=\angle A C B \).
 Solution:

$AC$ is the diameter.

Angle in semicircle makes an angle $90^o$.

This implies,

$\angle ABC = 90^o$

In $\triangle ABC$,

$\angle CAB + \angle ABC + \angle ACB = 180^o$

$\angle CAB + \angle ACB = 180^o - 90^o = 90^o$.……….(i)

Diameter(radius) of a circle is perpendicular to the tangent.

Therefore,

$CA\ \perp\ AT$

$\angle CAT = 90^o$

$\angle CAB + \angle BAT = 90^o$…….(ii)

From equations (i) and (ii),

$\angle CAB + \angle ACB = \angle CAB + \angle BAT$

$\angle ACB = \angle BAT$

Hence proved.

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Updated on: 10-Oct-2022

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