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Question Number 179360 by a.lgnaoui last updated on 28/Oct/22

Donnes: AD=1; DB=6; ∡BCD=45° ; ∡BAC=90° Determiner 1) AC ? 2) AE? −−−−−−−−−−−− Solution △BAC ∡BAC=90° BC^2 =AB^2 +AC^2 CD (cote commun aux △BAC et BDC) DB^2 =CD^2 +BC^2 −2BC×CDcos 45° (1) △CAD CD^2 =AD^2 +AC^2 (1) DB^2 =(AD^2 +AC^2 )+(AB^2 +AC^2 )−2(√((AB^2 +AC^2 )(AD^2 +AC^2 ))) cos 45° DB^2 =(AD^2 +AB^2 +2AC^2 −(√(2(AB^2 +AC^2 )(AD^2 +AC^2 ))) 2(AB^2 +AC^2 )(AD^2 +AC^2 )=(AD^2 +AB^2 +2AC^2 −DB^2 )^2 posons: AC=x 2(49+x^2 )(1+x^2 )=(1+49+2x^2 −36)^2 (x^4 +50x^2 +49)=2(x^4 +14x^2 +49) x^4 −22x^2 +49=0 x^2 =11±6(√2) x=(√(11+6(√2) )) AC =4,4142135 2)AB et AC coupent le cercle en (D,B) et( E,C) Nous avons AD×AB=AE×AC AE=((AD×AB)/(AC))=(7/( (√(11+6(√2)))))=((7(√(11+6(√2))))/(11+6(√2))) AE=1,585786

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

Answered by a.lgnaoui last updated on 28/Oct/22

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