Question

prove that r1(r2+r3)/√r1r2+r2r3+r3r1=a​

Answer 1

Properties of Triangles.

If $‘r’$ is radius of in circle and $r_{1},$ $r_{2},$ $r_{3}$ are the radii of ex-circles opposite to the vertices $A,$ $B,$ $C$ of $\Delta ABC$ respectively then

$r=\dfrac{\Delta}{s},\:r_{1}=\dfrac{\Delta}{s-a},\:r_{2}=\dfrac{\Delta}{s-b},\:r_{3}=\dfrac{\Delta}{s-c}$

where $2s=a+b+c$

To prove.

$\dfrac{r_{1}(r_{2}+r_{3})}{\sqrt{r_{1}r_{2}+r_{2}r_{3}+r_{3}r_{1}}}=a$

Step-by-step explanation.

Now $\dfrac{r_{1}(r_{2}+r_{3})}{\sqrt{r_{1}r_{2}+r_{2}r_{3}+r_{3}r_{1}}}$

Put values $r_{1}=\dfrac{\Delta}{s-a},\: r_{2}=\dfrac{\Delta}{r_{2}},\: r_{3}=\dfrac{\Delta}{s-c}$

$=\dfrac{\dfrac{\Delta}{s-a}(\dfrac{\Delta}{s-b}+\dfrac{\Delta}{s-c})}{\sqrt{\dfrac{\Delta}{s-a}\dfrac{\Delta}{s-b}+\dfrac{\Delta}{s-b}\dfrac{\Delta}{s-c}+\dfrac{\Delta}{s-c}\dfrac{\Delta}{s-a}}}$

Take $\Delta$ or $\Delta^{2}$ common from respective terms

$=\dfrac{\Delta^{2}.\dfrac{1}{s-a}(\dfrac{1}{s-b}+\dfrac{1}{s-c})}{\sqrt{\Delta^{2}(\dfrac{1}{(s-a)(s-b)}+\dfrac{1}{(s-b)(s-c)}+\dfrac{1}{(s-c)(s-a)})}}$

Do LCM and use the law of indices $\sqrt{a^{2}}=a$

$=\dfrac{\dfrac{\Delta^{2}}{s-a}.\dfrac{s-c+s-b}{(s-b)(s-c)}}{\Delta\sqrt{\dfrac{s-c+s-a+s-b}{(s-a)(s-b)(s-c}}}$

Divide both the numerator and the denominator by $\Delta$ where $\Delta\neq 0$

$=\dfrac{\dfrac{\Delta}{s-a}.\dfrac{2s-b-c}{(s-b)(s-c)}}{\sqrt{\dfrac{3s-a-b-c}{(s-a)(s-b)(s-c)}}}$

Use property of triangle $2s=a+b+c$

$=\dfrac{\dfrac{\Delta\:a}{(s-a)(s-b)(s-c)}}{\sqrt{\dfrac{s}{(s-a)(s-b)(s-c)}}}$

Use the ratio property $\dfrac{\dfrac{a}{b}}{\dfrac{c}{d}}=\dfrac{a}{b}.\dfrac{d}{c}$

$=\dfrac{\Delta\:a}{(s-a)(s-b)(s-c)}.\sqrt{\dfrac{(s-a)(s-b)(s-c)}{s}}$

Use laws of indices for $a=\sqrt{a^{2}}$ and $\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}$

$=\dfrac{\Delta\:a}{\sqrt{\{(s-a)(s-b)(s-c)\}^{2}}}.\dfrac{\sqrt{(s-a)(s-b)(s-c)}}{\sqrt{s}}$

Use laws of indices $\dfrac{a}{a^{2}}=\dfrac{1}{a^{2-1}}=\dfrac{1}{a}$ and $\sqrt{a}.\sqrt{b}=\sqrt{ab}$

$=\dfrac{\Delta\:a}{\sqrt{s(s-a)(s-b)(s-c)}}$

Use property of triangles $\Delta=\sqrt{s(s-a)(s-b)(s-c)}$

$=\frac{\Delta\:a}{\Delta}$

Since $\Delta\neq 0,$ we divide both the numerator and the denominator by $\Delta$

$=a$

Conclusion.

$\dfrac{r_{1}(r_{2}+r_{3})}{\sqrt{r_{1}r_{2}+r_{2}r_{3}+r_{3}r_{1}}}=a$

Thus proved.