Question Number 179360 by a.lgnaoui last updated on 28/Oct/22
$${Donnes}:\:\mathrm{AD}=\mathrm{1};\:\mathrm{DB}=\mathrm{6};\:\measuredangle{BCD}=\mathrm{45}°\:;\:\measuredangle{BAC}=\mathrm{90}° \\ $$$${Determiner} \\ $$$$\left.\mathrm{1}\right)\:\:\mathrm{A}{C}\:\:?\:\: \\ $$$$\left.\mathrm{2}\right)\:\:\mathrm{A}{E}? \\ $$$$−−−−−−−−−−−− \\ $$$${Solution} \\ $$$$\bigtriangleup{BAC}\:\:\:\:\:\:\measuredangle{BAC}=\mathrm{90}° \\ $$$${BC}^{\mathrm{2}} ={AB}^{\mathrm{2}} +{AC}^{\mathrm{2}} \\ $$$${CD}\:\left({cote}\:{commun}\:{aux}\:\bigtriangleup{BAC}\:{et}\:{BDC}\right) \\ $$$${DB}^{\mathrm{2}} ={CD}^{\mathrm{2}} +{BC}^{\mathrm{2}} −\mathrm{2}{BC}×{CD}\mathrm{cos}\:\mathrm{45}°\:\:\left(\mathrm{1}\right) \\ $$$$\bigtriangleup{CAD}\:\:\:{CD}^{\mathrm{2}} ={AD}^{\mathrm{2}} +{AC}^{\mathrm{2}} \\ $$$$\left(\mathrm{1}\right)\:{DB}^{\mathrm{2}} =\left({AD}^{\mathrm{2}} +{AC}^{\mathrm{2}} \right)+\left({AB}^{\mathrm{2}} +{AC}^{\mathrm{2}} \right)−\mathrm{2}\sqrt{\left({AB}^{\mathrm{2}} +{AC}^{\mathrm{2}} \right)\left({AD}^{\mathrm{2}} +{AC}^{\mathrm{2}} \right)}\:\mathrm{cos}\:\mathrm{45}° \\ $$$$\:\:{DB}^{\mathrm{2}} =\left({AD}^{\mathrm{2}} +{AB}^{\mathrm{2}} +\mathrm{2}{AC}^{\mathrm{2}} −\sqrt{\mathrm{2}\left({AB}^{\mathrm{2}} +{AC}^{\mathrm{2}} \right)\left({AD}^{\mathrm{2}} +{AC}^{\mathrm{2}} \right)}\:\:\right. \\ $$$$\mathrm{2}\left({AB}^{\mathrm{2}} +{AC}^{\mathrm{2}} \right)\left({AD}^{\mathrm{2}} +{AC}^{\mathrm{2}} \right)=\left({AD}^{\mathrm{2}} +{AB}^{\mathrm{2}} +\mathrm{2}{AC}^{\mathrm{2}} −{DB}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$${posons}:\:\:\:{AC}={x} \\ $$$$\mathrm{2}\left(\mathrm{49}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)=\left(\mathrm{1}+\mathrm{49}+\mathrm{2}{x}^{\mathrm{2}} −\mathrm{36}\right)^{\mathrm{2}} \\ $$$$\left({x}^{\mathrm{4}} +\mathrm{50}{x}^{\mathrm{2}} +\mathrm{49}\right)=\mathrm{2}\left({x}^{\mathrm{4}} +\mathrm{14}{x}^{\mathrm{2}} +\mathrm{49}\right) \\ $$$$\:\:\:{x}^{\mathrm{4}} −\mathrm{22}{x}^{\mathrm{2}} +\mathrm{49}=\mathrm{0} \\ $$$$\:{x}^{\mathrm{2}} =\mathrm{11}\pm\mathrm{6}\sqrt{\mathrm{2}}\:\:\:\:{x}=\sqrt{\mathrm{11}+\mathrm{6}\sqrt{\mathrm{2}}\:}\: \\ $$$$\:{AC}\:=\mathrm{4},\mathrm{4142135} \\ $$$$\left.\:\mathrm{2}\right){AB}\:{et}\:\:{AC}\:{coupent}\:{le}\:{cercle}\:\:{en}\:\left({D},{B}\right)\:{et}\left(\:{E},{C}\right) \\ $$$$\:{Nous}\:{avons}\:\:{AD}×{AB}={AE}×{AC} \\ $$$$\:\:{AE}=\frac{{AD}×{AB}}{{AC}}=\frac{\mathrm{7}}{\:\sqrt{\mathrm{11}+\mathrm{6}\sqrt{\mathrm{2}}}}=\frac{\mathrm{7}\sqrt{\mathrm{11}+\mathrm{6}\sqrt{\mathrm{2}}}}{\mathrm{11}+\mathrm{6}\sqrt{\mathrm{2}}} \\ $$$$\:{AE}=\mathrm{1},\mathrm{585786} \\ $$$$ \\ $$
Answered by a.lgnaoui last updated on 28/Oct/22
