# In a triangle ABC prove this :"

Dear Student,
Please find below the solution to the asked query:

$\mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{1}{{\mathrm{r}}^{2}}+\frac{1}{{{\mathrm{r}}_{1}}^{2}}+\frac{1}{{{\mathrm{r}}_{2}}^{2}}+\frac{1}{{{\mathrm{r}}_{3}}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{1}{\frac{{∆}^{2}}{{\mathrm{s}}^{2}}}+\frac{1}{\frac{{∆}^{2}}{{\left(\mathrm{s}-\mathrm{a}\right)}^{2}}}+\frac{1}{\frac{{∆}^{2}}{{\left(\mathrm{s}-\mathrm{b}\right)}^{2}}}+\frac{1}{\frac{{∆}^{2}}{{\left(\mathrm{s}-\mathrm{c}\right)}^{2}}}\phantom{\rule{0ex}{0ex}}=\frac{1}{{∆}^{2}}\left\{{\mathrm{s}}^{2}+{\left(\mathrm{s}-\mathrm{a}\right)}^{2}+{\left(\mathrm{s}-\mathrm{b}\right)}^{2}+{\left(\mathrm{s}-\mathrm{c}\right)}^{2}\right\}\phantom{\rule{0ex}{0ex}}=\frac{1}{{∆}^{2}}\left\{{\mathrm{s}}^{2}+{\mathrm{s}}^{2}+{\mathrm{a}}^{2}-2\mathrm{as}+{\mathrm{s}}^{2}+{\mathrm{b}}^{2}-2\mathrm{bs}+{\mathrm{s}}^{2}+{\mathrm{c}}^{2}-2\mathrm{cs}\right\}\phantom{\rule{0ex}{0ex}}=\frac{1}{{∆}^{2}}\left\{4{\mathrm{s}}^{2}+{\mathrm{a}}^{2}+{\mathrm{b}}^{2}+{\mathrm{c}}^{2}-2\mathrm{s}\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)\right\}\phantom{\rule{0ex}{0ex}}=\frac{1}{{∆}^{2}}\left\{4{\left(\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{2}\right)}^{2}+{\mathrm{a}}^{2}+{\mathrm{b}}^{2}+{\mathrm{c}}^{2}-2\left(\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{2}\right)\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)\right\}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{1}{{∆}^{2}}\left\{4\frac{{\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)}^{2}}{4}+{\mathrm{a}}^{2}+{\mathrm{b}}^{2}+{\mathrm{c}}^{2}-{\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)}^{2}\right\}\phantom{\rule{0ex}{0ex}}=\frac{1}{{∆}^{2}}\left\{{\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)}^{2}+{\mathrm{a}}^{2}+{\mathrm{b}}^{2}+{\mathrm{c}}^{2}-{\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)}^{2}\right\}\phantom{\rule{0ex}{0ex}}=\frac{1}{{∆}^{2}}\left({\mathrm{a}}^{2}+{\mathrm{b}}^{2}+{\mathrm{c}}^{2}\right)\phantom{\rule{0ex}{0ex}}=\frac{16{\mathrm{a}}^{2}{\mathrm{b}}^{2}{\mathrm{c}}^{2}}{{∆}^{2}}\left(\frac{{\mathrm{a}}^{2}+{\mathrm{b}}^{2}+{\mathrm{c}}^{2}}{16{\mathrm{a}}^{2}{\mathrm{b}}^{2}{\mathrm{c}}^{2}}\right)\phantom{\rule{0ex}{0ex}}=16{\left(\frac{\mathrm{abc}}{4∆}\right)}^{2}\left(\frac{{\mathrm{a}}^{2}+{\mathrm{b}}^{2}+{\mathrm{c}}^{2}}{{\mathrm{a}}^{2}{\mathrm{b}}^{2}{\mathrm{c}}^{2}}\right)\phantom{\rule{0ex}{0ex}}=16{\mathrm{R}}^{2}\left(\frac{{\mathrm{a}}^{2}+{\mathrm{b}}^{2}+{\mathrm{c}}^{2}}{{\mathrm{a}}^{2}{\mathrm{b}}^{2}{\mathrm{c}}^{2}}\right)\phantom{\rule{0ex}{0ex}}\mathrm{On}\mathrm{comparing},\mathrm{we}\mathrm{get}:\phantom{\rule{0ex}{0ex}}\mathbf{k}\mathbf{=}\mathbf{16}\mathbf{}\left(\mathbf{Answer}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$