Question Number 226770 by gregori last updated on 14/Dec/25
Answered by TonyCWX last updated on 14/Dec/25
$$\mathrm{Stewart}'\mathrm{s}\:\mathrm{Theorem}: \\ $$$${b}^{\mathrm{2}} {x}+{b}^{\mathrm{2}} {y}=\left({x}+{y}\right)\left({z}^{\mathrm{2}} −{xy}\right) \\ $$$${b}^{\mathrm{2}} \left({x}+{y}\right)=\left({x}+{y}\right)\left({z}^{\mathrm{2}} −{xy}\right) \\ $$$${z}^{\mathrm{2}} −{xy}={b}^{\mathrm{2}} \\ $$$${xy}={b}^{\mathrm{2}} −{z}^{\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{Pythagoras}'\:\mathrm{Theorem}: \\ $$$$\left({x}+{y}\right)^{\mathrm{2}} =\mathrm{2}{b}^{\mathrm{2}} \\ $$$$ \\ $$$$\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{{z}^{\mathrm{2}} }=\frac{\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{2}{xy}}{{z}^{\mathrm{2}} }=\frac{\mathrm{2}{b}^{\mathrm{2}} −\mathrm{2}\left({b}^{\mathrm{2}} −{z}^{\mathrm{2}} \right)}{{z}^{\mathrm{2}} }=\frac{\mathrm{2}{z}^{\mathrm{2}} }{{z}^{\mathrm{2}} }=\mathrm{2} \\ $$
Answered by mr W last updated on 14/Dec/25
$$\sqrt{\mathrm{2}}{b}={x}+{y},\:\mathrm{2}{b}^{\mathrm{2}} =\left({x}+{y}\right)^{\mathrm{2}} \\ $$$${z}^{\mathrm{2}} ={x}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{bx}\:\mathrm{cos}\:\mathrm{45}° \\ $$$$\:\:\:\:={x}^{\mathrm{2}} +{b}^{\mathrm{2}} −\sqrt{\mathrm{2}}{bx} \\ $$$$\:\:\:\:={x}^{\mathrm{2}} +\frac{\left({x}+{y}\right)^{\mathrm{2}} }{\mathrm{2}}−\left({x}+{y}\right){x} \\ $$$$\:\:\:\:={x}^{\mathrm{2}} +\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{2}{xy}}{\mathrm{2}}−{x}^{\mathrm{2}} −{xy} \\ $$$$\:\:\:\:=\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\Rightarrow\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{{z}^{\mathrm{2}} }=\mathrm{2}\:\checkmark \\ $$