Question

In a right angled triangle $ABC$, $\angle A={90}^{\circ }$ and $D$ is the midpoint of $BC$. If $r_1,r_2$ are the inradii of $△ABD$ and $△ACD$ respectively, then which of the following is/are correct ?

• A. $\frac{r_1}{r_2}{|}_{max}\to 4$
• B. $\frac{r_1}{r_2}{|}_{min}\to \frac{1}{2}$
• C. $\frac{r_1}{r_2}{|}_{min}\to \frac{1}{4}$
• D. $\frac{r_1}{r_2}{|}_{max}\to 2$

Answer 1

Answer: (B), (D)

The correct options are
B $\frac{r_1}{r_2}{|}_{min}\to \frac{1}{2}$
D $\frac{r_1}{r_2}{|}_{max}\to 2$


Let height of the triangles $△ABD$ and $△ACD$ through $A$ be $h$
$BD$ and $AD$ are circumradius of $△ABC$
$\therefore AD=BD=CD=\frac{a}{2}$

In $△ABD,$
$r_1=\frac{{\Delta }_1}{s_1}$
$=\frac{\frac{1}{2}\cdot \left(\frac{a}{2}\right)\cdot h}{\frac{\frac{a}{2}+\frac{a}{2}+c}{2}}$
$=\frac{ah}{2(a+c)}$

In $△ACD,$
$r_2=\frac{{\Delta }_2}{s_2}$
$=\frac{\frac{1}{2}\cdot \left(\frac{a}{2}\right)\cdot h}{\frac{\frac{a}{2}+\frac{a}{2}+b}{2}}$
$=\frac{ah}{2(a+b)}$

$\Rightarrow \frac{r_1}{r_2}=\frac{a+b}{a+c}$

$=\frac{2R\sin A+2R\sin B}{2R\sin A+2R\sin C}$

$=\frac{1+\sin B}{1+\sin C}(\because A={90}^{\circ })$

$=\frac{1+\sin B}{1+\cos B}(\because C={90}^{\circ }-B)$

$\frac{r_1}{r_2}{|}_{max}\to 2$ when $\angle B\to {90}^{\circ }$

$\frac{r_1}{r_2}{|}_{min}\to \frac{1}{2}$ when $\angle B\to 0^{\circ }$