Question

Question 4
If AB is a chord of a circle with centre O, AOC is a diameter and tangent T at A as shown in the figure. Prove that $\angle$BAT = $\angle$ACB.

Answer 1

AC is a diameter.
We know that angle in a semi-circle is ${90}^{\circ }$.
$\therefore \angle ABC={90}^{\circ }$ [ by property]
In $\Delta$ ABC. $\angle CAB+\angle ABC+\angle ACB={180}^{\circ }$
[ $\because$ sum of all interior angles of any triangle is ${180}^{\circ }$]
$\Rightarrow \angle CAB+\angle ACB={180}^{\circ }-{90}^{\circ }={90}^{\circ }$ ...(i)
Since , diameter of a circle is perpendicular to the tangent.
i.e CA $\perp$ AT
$\therefore \angle CAT={90}^{\circ }$
$\Rightarrow \angle CAB+\angle BAT={90}^{\circ }$ ...(ii)
From Eqs. (i) and (ii)
$\angle CAB+\angle ACB=\angle CAB+\angle BAT$
$\Rightarrow \angle ACB=\angle BAT$
Hence proved.