In Fig. 4, $AB$ is a chord of circle with centre $O$, $AOC$ is diameter and $AT$ is tangent at $A$. Prove that $\angle BAT = \angle ACB$.

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Question

In Fig. 4, $AB$ is a chord of circle with centre $O$, $AOC$ is diameter and $AT$ is tangent at $A$. Prove that $\angle BAT = \angle ACB$.

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Accepted Answer

Step 1: $\angle BAC = 90° - \angle BAT$ ... (i). In $	riangle BAC$, $\angle B = 90°$ (angle in semicircle)
Step 2: $	herefore \angle BCA = 90° - \angle BAC$. Or $\angle ACB = \angle BAT$ (Using (i))

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