Question Number 198059 by a.lgnaoui last updated on 09/Oct/23
$$\mathrm{Montrer}\:\mathrm{l}\:\mathrm{egalite}\:\:\left(\mathrm{voir}\:\mathrm{la}\:\mathrm{figure}\:\right) \\ $$
Commented by a.lgnaoui last updated on 09/Oct/23
Answered by mr W last updated on 10/Oct/23
$${a}^{\mathrm{2}} ={z}^{\mathrm{2}} +{y}^{\mathrm{2}} \\ $$$${b}^{\mathrm{2}} ={x}^{\mathrm{2}} +{z}^{\mathrm{2}} \\ $$$${c}^{\mathrm{2}} ={y}^{\mathrm{2}} +{x}^{\mathrm{2}} \\ $$$$\Delta{ABC}=\frac{\sqrt{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2}\left({a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} \right)}}{\mathrm{4}} \\ $$$$\Delta{ABC}^{\mathrm{2}} =\frac{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2}\left({a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} \right)}{\mathrm{16}} \\ $$$$\Delta{ABC}^{\mathrm{2}} =\frac{\mathrm{4}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2}\left({x}^{\mathrm{4}} +{y}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{2}} {y}^{\mathrm{2}} +{y}^{\mathrm{4}} +{z}^{\mathrm{4}} +\mathrm{2}{y}^{\mathrm{2}} {z}^{\mathrm{2}} +{z}^{\mathrm{4}} +{x}^{\mathrm{4}} +\mathrm{2}{z}^{\mathrm{2}} {x}^{\mathrm{2}} \right)}{\mathrm{16}} \\ $$$$\Delta{ABC}^{\mathrm{2}} =\frac{\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)^{\mathrm{2}} −\left({x}^{\mathrm{4}} +{y}^{\mathrm{4}} +{z}^{\mathrm{4}} +{x}^{\mathrm{2}} {y}^{\mathrm{2}} +{y}^{\mathrm{2}} {z}^{\mathrm{2}} +{z}^{\mathrm{2}} {x}^{\mathrm{2}} \right)}{\mathrm{4}} \\ $$$$\Delta{ABC}^{\mathrm{2}} =\frac{{x}^{\mathrm{2}} {y}^{\mathrm{2}} +{y}^{\mathrm{2}} {z}^{\mathrm{2}} +{z}^{\mathrm{2}} {x}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\Delta{ABC}^{\mathrm{2}} =\left(\frac{{xy}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{{yz}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{{zx}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\Delta{ABC}^{\mathrm{2}} =\Delta{ABD}^{\mathrm{2}} +\Delta{BCD}^{\mathrm{2}} +\Delta{ACD}^{\mathrm{2}} \:\checkmark \\ $$
Commented by a.lgnaoui last updated on 09/Oct/23
$$\mathrm{thanks}\: \\ $$
Commented by a.lgnaoui last updated on 09/Oct/23
$$\mathrm{remarque}:\:\mathrm{a}^{\mathrm{2}} =\mathrm{y}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} \:\:\:\:\mathrm{b}^{\mathrm{2}} =\mathrm{x}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{c}^{\mathrm{2}} =\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}.} \\ $$