Question

The radius of the circle passing through the center of incircle of $ABC$ and through the end points of $BC$ is given by

• A. $\frac{a}{2}$
• B. $\frac{a}{2}\sec \frac{A}{2}$
• C. $\frac{a}{2}\sin A$
• D. $a\sec \frac{A}{2}$

Answer 1

The correct option is B $\frac{a}{2}\sec \frac{A}{2}$

$\angle BIC+\angle BDC={180}^0$

$\therefore \angle BDC={90}^0-\frac{A}{2}(\because \angle BIC={90}^0+\frac{A}{2})$

Let $O$ be the center of the required circle and $R_1$ is the radius.

Then $\angle BOC=2\angle BDC=({180}^0-A);BC=a$

In $△BOC,$

$\cos ({180}^0-A)=\frac{{R_1}^2+{R_1}^2-a^2}{2R_1{R}_1}$

$\Rightarrow -2{R_1}^2\cos A=2{R_1}^2-a^2$

$\Rightarrow a^2=2{R_1}^2(1+\cos A)\Rightarrow a^2=4{R_1}^2{\cos }^2\frac{A}{2}$

$\Rightarrow {R_1}^2=\frac{a^2}{4}{\sec }^2\frac{A}{2}\Rightarrow R_1=\frac{a}{2}\sec \frac{A}{2}$