| 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
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