Question Number 148932 by vvvv last updated on 01/Aug/21
$$\frac{\left(\boldsymbol{{b}}+\boldsymbol{{c}}\right)^{\mathrm{2}} }{\boldsymbol{{bc}}}\boldsymbol{{l}}_{\boldsymbol{{a}}} ^{\mathrm{2}} +\frac{\left(\boldsymbol{{a}}+\boldsymbol{{b}}\right)^{\mathrm{2}} }{\boldsymbol{{ab}}}\boldsymbol{{l}}_{\boldsymbol{{c}}} ^{\mathrm{2}} +\frac{\left(\boldsymbol{{a}}+\boldsymbol{{c}}\right)^{\mathrm{2}} }{\boldsymbol{{ac}}}\boldsymbol{{l}}_{\boldsymbol{{b}}} ^{\mathrm{2}} =\left(\boldsymbol{{a}}+\boldsymbol{{b}}+\boldsymbol{{c}}\right)^{\mathrm{2}} \\ $$$$\boldsymbol{{l}}_{\boldsymbol{{b}}} ,\boldsymbol{{l}}_{\boldsymbol{{a}}} ,\boldsymbol{{l}}_{\boldsymbol{{c}}} −\boldsymbol{{bissekterissa}} \\ $$$$\boldsymbol{{prove}} \\ $$
Answered by mr W last updated on 01/Aug/21
$${s}=\frac{{a}+{b}+{c}}{\mathrm{2}} \\ $$$${l}_{{a}} =\frac{\mathrm{2}}{{b}+{c}}\sqrt{{bcs}\left({s}−{a}\right)} \\ $$$$\frac{\left({b}+{c}\right)^{\mathrm{2}} }{{bc}}{l}_{{a}} ^{\mathrm{2}} =\frac{\left({b}+{c}\right)^{\mathrm{2}} }{{bc}}×\frac{\mathrm{4}{bcs}\left({s}−{a}\right)}{\left({b}+{c}\right)^{\mathrm{2}} }=\mathrm{4}{s}\left({s}−{a}\right) \\ $$$$\Sigma\frac{\left({b}+{c}\right)^{\mathrm{2}} }{{bc}}{l}_{{a}} ^{\mathrm{2}} =\mathrm{4}{s}\left(\mathrm{3}{s}−{a}−{b}−{c}\right)=\mathrm{4}{s}^{\mathrm{2}} \\ $$$$=\left({a}+{b}+{c}\right)^{\mathrm{2}} \\ $$
Commented by vvvv last updated on 01/Aug/21
$$\boldsymbol{{thank}}\:\boldsymbol{{you}} \\ $$