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 Fig. 9.17. Prove that
\(\angle BAT = \angle ACB\).

\(\angle BAT = \angle ACB\)
Step 1: Look at the figure. AB is a chord of the circle, AOC is a diameter, and AT is a tangent at point A.
Step 2: Recall the tangent–chord theorem (also called the Alternate Segment Theorem). It says: The angle between a tangent and a chord through the point of contact is equal to the angle subtended by the chord in the alternate segment of the circle.
Step 3: Here, the tangent is AT, and the chord through the point of contact is AB.
Step 4: So, the angle between AT and AB is \(\angle BAT\).
Step 5: According to the theorem, this angle must be equal to the angle subtended by the same chord AB in the alternate segment of the circle.
Step 6: The chord AB subtends an angle at point C on the opposite segment of the circle, which is \(\angle ACB\).
Step 7: Therefore, by the tangent–chord theorem, we get: \[ \angle BAT = \angle ACB \]
Final Result: Hence proved that \(\angle BAT = \angle ACB\).