Question

DEF is the triangle formed by joining the points of contact of the incircle with the sides of the $\Delta ABC$, prove that its sides are 2r $\cos \frac{A}{2},$ 2r $\cos \frac{B}{2},$ and 2r 2r $\cos \frac{C}{2},$ where r is the radius of incircle of $\Delta ABC.$ and its angles are $\left(\frac{\pi }{2}-\frac{A}{2}\right),(\frac{\pi }{2}-\frac{B}{2})$ and $\left(\frac{\pi }{2}-\frac{C}{2}\right)$. It's area is $\frac{r^{2}s}{2R}$ where 's' is the semiperimeter and R is the circumradius of the $\Delta ABC$.

• A. $\frac{r^{2}s}{2R}.$
• B. $\frac{r^{2}s}{R}.$
• C. $\frac{R^{2}s}{2r}.$
• D. $\frac{R^{2}s}{r}.$

Answer 1

The correct option is A $\frac{r^{2}s}{2R}.$

Applying sine rule in $\Delta DBF\frac{r.\cot B/2}{\cos B/2}=\frac{DF}{\sin B}$ 2r $\cos \frac{B}{2}=DF$
Similarly $2r\cos \frac{C}{2}=EF$ and $2r\cos \frac{A}{2}=DE$

$\angle BDF=\angle BFD=\frac{\pi -B}{2}\Rightarrow \angle DFI=\frac{A}{2}$
Similarly $\angle IFE=\frac{C}{2}$

$\therefore \angle DFE=\frac{B+C}{2}=\frac{\pi }{2}-\frac{A}{2}$

Similarly $DEF=\frac{\pi }{2}-\frac{C}{2}$ and $\angle EDF=\frac{\pi }{2}-\frac{C}{2}$

Area $\left(\Delta DEF\right)=2r^2\cos \frac{B}{2}.\cos \frac{C}{2}.\cos \frac{A}{2}$

$=2r^2\sqrt{\frac{\left(s\right)(s-b)}{ac}}\sqrt{\frac{s(s-c)}{ab}}\sqrt{\frac{\left(s\right)(s-a)}{bc}}$
$=\frac{2rs\Delta }{abc}=\frac{r^{2}s}{2R}.$