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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

$$a^2=z^2+y^2$$$$b^2=x^2+z^2$$$$c^2=y^2+x^2$$$$\mathrm{\Delta }ABC=\frac{\sqrt{{(a^2+b^2+c^2)}^2-2(a^4+b^4+c^4)}}{4}$$$$\mathrm{\Delta }{ABC}^2=\frac{{(a^2+b^2+c^2)}^2-2(a^4+b^4+c^4)}{16}$$$$\mathrm{\Delta }{ABC}^2=\frac{4{(x^2+y^2+z^2)}^2-2(x^4+y^4+2x^2{y}^2+y^4+z^4+2y^2{z}^2+z^4+x^4+2z^2{x}^2)}{16}$$$$\mathrm{\Delta }{ABC}^2=\frac{{(x^2+y^2+z^2)}^2-(x^4+y^4+z^4+x^2{y}^2+y^2{z}^2+z^2{x}^2)}{4}$$$$\mathrm{\Delta }{ABC}^2=\frac{x^2{y}^2+y^2{z}^2+z^2{x}^2}{4}$$$$\mathrm{\Delta }{ABC}^2={\left(\frac{xy}{2}\right)}^2+{\left(\frac{yz}{2}\right)}^2+{\left(\frac{zx}{2}\right)}^2$$$$\Rightarrow \mathrm{\Delta }{ABC}^2=\mathrm{\Delta }{ABD}^2+\mathrm{\Delta }{BCD}^2+\mathrm{\Delta }{ACD}^2✓$$

Commented by a.lgnaoui last updated on 09/Oct/23

$$\mathrm{thanks}$$

Commented by a.lgnaoui last updated on 09/Oct/23

$$\mathrm{remarque}:{\mathrm{a}}^2={\mathrm{y}}^2+{\mathrm{z}}^2{\mathrm{b}}^2={\mathrm{x}}^2+{\mathrm{z}}^2$$$${\mathrm{c}}^2={\mathrm{x}}^2+{\mathrm{y}}^{2.}$$

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