Question

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

• A. $\frac{a}{2}\cos A$
• 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 $\frac{a}{2}\sec \frac{A}{2}$

From the figure:
In $\Delta BIC$: $\angle BIC=\pi -\left(\frac{B+C}{2}\right)$
From sine rule: $\frac{a}{sin\angle BIC}=2R_1$
where $R_1$ is cricumradius of $\Delta BIC$
$\Rightarrow R_1=\frac{a}{2sin(\pi -\frac{B+C}{2})}=\frac{a}{2}\sec \frac{A}{2}$
Ans: B