Question Number 209220 by alcohol last updated on 04/Jul/24
Answered by Berbere last updated on 04/Jul/24
$${SAB}\:,{SAC}\:\:\&{ABC}\:{c}\:{est}\:{claire}\:\left({AS}\right)\bot\left({ABC}\right)\:\&{ABC}\:{rectangle} \\ $$$$\Rightarrow{SA}^{\mathrm{2}} +{AB}^{\mathrm{2}} ={SB}^{\mathrm{2}} ;{SA}^{\mathrm{2}} +{AC}^{\mathrm{2}} ={SC}^{\mathrm{2}} ;{CA}^{\mathrm{2}} +{CB}^{\mathrm{2}} ={AB}^{\mathrm{2}} \\ $$$$\left(\mathrm{1}\right)\:\&\left(\mathrm{2}\right)\Rightarrow{SC}^{\mathrm{2}} −{SB}^{\mathrm{2}} ={AC}^{\mathrm{2}} −{AB}^{\mathrm{2}} \\ $$$$\Rightarrow{SB}^{\mathrm{1}} −{SC}^{\mathrm{2}} ={CB}^{\mathrm{2}} \Rightarrow{SB}^{\mathrm{2}} ={CB}^{\mathrm{2}} +{SC}^{\mathrm{2}} \Rightarrow{SBC}\:\:{est}\:{rectangle}\:{en}\:{C} \\ $$$$\left(\mathrm{2}\right) \\ $$$${AF}\:\bot\left({SB}\right)\: \\ $$$${A}\overset{\rightarrow} {{F}}.{C}\overset{\rightarrow} {{B}}=\left({A}\overset{\rightarrow} {{C}}+\overset{\rightarrow} {{C}F}\right).{C}\overset{\rightarrow} {{B}}={A}\overset{\rightarrow} {{C}}.{C}\overset{\rightarrow} {{B}}+{C}\overset{\rightarrow} {{F}}.{C}\overset{\rightarrow} {{B}}=\mathrm{0}+\mathrm{0}=\mathrm{0} \\ $$$${first}\:{ACB}\:{rectangle}\:\left(\mathrm{2}\right);{SCB}\:{rectangle}\:{C}\overset{\rightarrow} {{F}}={aS}\overset{\rightarrow} {{C}} \\ $$$$\left({AF}\right)\bot\left({CB}\right) \\ $$$$\left(\mathrm{2}\right){bAF}\in\left({SAC}\right)\:{and}\:\left({AF}\right)\bot\left({SBC}\right)\Rightarrow\left({ASC}\right)\bot\left({SBC}\right) \\ $$$$\left({AC}\right)\in\left({SAC}\right)\:\&\:\left({AC}\right)\bot\left({BC}\right);\left({AC}\right)\bot\left({SC}\right) \\ $$$$\Rightarrow\left({AC}\right)\in\left({ASC}\right)\:\&\left({AC}\right)\bot\left({SBC}\right)\Rightarrow\left({ASC}\right)\bot\left({SBC}\right) \\ $$$$\left(\mathrm{3}\right) \\ $$$$\left({AF}\right)\in\left({SAC}\right)\bot\left({SBC}\right) \\ $$$$\Rightarrow\left({AF}\right)\bot\left({BC}\right) \\ $$$${A}\overset{\rightarrow} {{F}}={A}\overset{\rightarrow} {{S}}+{aS}\overset{\rightarrow} {{C}};{A}\overset{\rightarrow} {{F}}.{A}\overset{\rightarrow} {{C}}=\mathrm{0}\:{car}\:{les}\:{face}\:{du}\:{Tetraede}\:{Sont} \\ $$$${orthogonal} \\ $$$$\left({AF}\right)\bot\left({AC}\right)\Rightarrow\left({AF}\right)\bot\left({ABC}\right)\Rightarrow\left({CFA}\right)\bot\left({ABC}\right) \\ $$$$\left({AF}\right)\bot\left({SBC}\right)\:\:;\left({AF}\right)\in\left({ADF}\right)\Rightarrow\left({ADF}\right)\bot\left({SBC}\right) \\ $$$${DF}\in\left({ADF}\right);\left({SB}\right)\in\left({SBC}\right)\Rightarrow\left({DF}\right)\bot\left({SB}\right) \\ $$$$\left(\mathrm{5}\right)\:{meme}\:{methode} \\ $$
Commented by alcohol last updated on 05/Jul/24
$${le}\:{schema}\:{pardon} \\ $$