Question

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. 

Answer 1

In the given figure, 
AC is the diameter.
So, 

\[\angle CBA = 90^o\](Angle formed by the diameter on the circle is 90º)
AT is the tangent at point A. 
Thus, 

\[\angle CAT = 90^o\]

In ∆ABC,

\[\angle BCA + \angle BAC + \angle CBA = 180^o \left( \text{Angle sum property} \right)\]
\[ \Rightarrow \angle BCA + 90^o + \angle CAT - \angle BAT = 180^o \]
\[ \Rightarrow \angle BCA + 90^o + 90 - \angle BAT = 180^o \]
\[ \Rightarrow \angle BCA = \angle BAT\]

Hence Proved