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*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 . 10.70. Prove that \[\angle\]*BAT* = \[\angle\] *ACB*.

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#### Solution

In the given figure,

AC is the diameter.

So,

AT is the tangent at point A.

Thus,

\[ \Rightarrow \angle BCA + 90^o + \angle CAT - \angle BAT = 180^o \]

\[ \Rightarrow \angle BCA + 90^o + 90 - \angle BAT = 180^o \]

\[ \Rightarrow \angle BCA = \angle BAT\]

Hence Proved

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