Question

If in a $△ABC,$ the incircle passing through the point of intersection of perpendicular bisector of sides $BC,AB.$ Then $4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}$ is equal to

• A. $\sqrt{2}$
• B. $\sqrt{2}-1$
• C. $\sqrt{2}+1$
• D. $\frac{1}{2}$

Answer 1

The correct option is B $\sqrt{2}-1$

Given that the circle passes through the circumcentre of $△ABC.$
Therefore, the distance between circumcentre, incentre
$=\sqrt{R^2-2rR}=r\Rightarrow r^2+2rR-R^2=0\Rightarrow {\left(\frac{r}{R}\right)}^2+2\frac{r}{R}-1=0\Rightarrow \frac{r}{R}=\sqrt{2}-1$
and $\cos A+\cos B+\cos C=1+\frac{r}{R}=1+4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}$
So, $4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}=\frac{r}{R}=\sqrt{2}-1$