Bonsoir :))
[tex]1.\ CHB\ est\ rectangle\ en\ H\\Donc\ BC^{2}=HB^{2}+HC^{2}\\AHC\ est\ rectangle\ en\ H\\Donc\ AC^{2}=HA^{2}+HC^{2}\\\\2.\ \Delta=AB^{2}+(HA^{2}+HC^{2})-(HB^{2}+HC^{2})\\\Delta=AB^{2}+(HA^{2}-HB^{2})\\\Delta=AB^{2}+(HA+HB)(HA-HB)\\\\3.AB=AH+HB\\\\4.\ \Delta=AB^{2}+AB(HA-HB)\\\Delta=AB(AB+(HA-HB))[/tex]
[tex]5.\ AH=AB-HB\\\\6.\ AB(AH+AH)\\\DElta=2\times AB\times AH\\\\7.\ Rappel\ trigonom\'etrie:\frac{C\^ot\'e\ adjacent}{hypot\'enuse}=\cos(\widehat{X})\\\\\frac{AH}{AC}=\cos(\widehat{BAC})\Leftrightarrow AH=AC\times \cos({BAC})\\\\8.\ \Delta=2\times AB\times AC\times \cos(\widehat{BAC})\\\\9.\ On\ sait\ que\ \Delta=AB^{2}+AC^{2}-BC^{2}\\Donc\ BC^{2}=AB^{2}+AC^{2}-\Delta\\\\\boxed{\bf{BC^{2}=AB^{2}+AC^{2}-2\times AB\times AC\times \cos(\widehat{BAC})}}[/tex]
Si besoin, tu peux me poser des questions!
Bonne continuation ! :))
