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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 Figure. Prove that \angleBAT = \angleACB

Answers (1)

Here    \angle ABC= 90^{\circ}     [ AC is a diameter line, \therefore Angle in semi circle formed, is 90^{\circ}]
   In \bigtriangleupABC
\angle CAB+\angle ABC+\angle BCA= 180^{\circ}
                      [\because  sum of interior angles of a triangle is 180^{\circ}]
\angle CAB+\angle BCA= 180^{\circ}-90^{\circ} ......(I)
We know that diameter of a circle is perpendicular to the tangent.
\therefore CA\perp AT
\therefore \angle CAT= 90^{\circ}
\Rightarrow \angle CAB+\angle BAT= 90^{\circ}\cdots \left ( ii \right )
Equate equation (i) and (ii) we get
\angle CAB+\angle BAT= \angle CAB+\angle BCA
\angle BAT= \angle BCA

Hence Proved

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