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Question Number 198059 by a.lgnaoui last updated on 09/Oct/23

Montrer l egalite (voir la figure )

$$\mathrm{Montrer}\:\mathrm{l}\:\mathrm{egalite}\:\:\left(\mathrm{voir}\:\mathrm{la}\:\mathrm{figure}\:\right) \\ $$

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 ΔABC=((√((a^2 +b^2 +c^2 )^2 −2(a^4 +b^4 +c^4 )))/4) ΔABC^2 =(((a^2 +b^2 +c^2 )^2 −2(a^4 +b^4 +c^4 ))/(16)) ΔABC^2 =((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)) ΔABC^2 =(((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) ΔABC^2 =((x^2 y^2 +y^2 z^2 +z^2 x^2 )/4) ΔABC^2 =(((xy)/2))^2 +(((yz)/2))^2 +(((zx)/2))^2 ⇒ΔABC^2 =ΔABD^2 +ΔBCD^2 +ΔACD^2 ✓

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

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

thanks

$$\mathrm{thanks}\: \\ $$

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

remarque: a^2 =y^2 +z^2 b^2 =x^2 +z^2 c^2 =x^2 +y^(2.)

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

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