如图,在△ABC中,AC=BC,∠ACB=90°,CD⊥AB,垂足为D,点E在AC上,BE交CD于点G,EF⊥BE交AB于点F,CE=kEA...如图,在△ABC中,AC=BC,∠ACB=90°,CD⊥AB,垂足为D,点E在AC上,BE交CD于点G,EF⊥BE交AB于点F,CE=kEA①若AC=BC,探究线段E
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![如图,在△ABC中,AC=BC,∠ACB=90°,CD⊥AB,垂足为D,点E在AC上,BE交CD于点G,EF⊥BE交AB于点F,CE=kEA...如图,在△ABC中,AC=BC,∠ACB=90°,CD⊥AB,垂足为D,点E在AC上,BE交CD于点G,EF⊥BE交AB于点F,CE=kEA①若AC=BC,探究线段E](/uploads/image/z/8656409-65-9.jpg?t=%E5%A6%82%E5%9B%BE%2C%E5%9C%A8%E2%96%B3ABC%E4%B8%AD%2CAC%3DBC%2C%E2%88%A0ACB%3D90%C2%B0%2CCD%E2%8A%A5AB%2C%E5%9E%82%E8%B6%B3%E4%B8%BAD%2C%E7%82%B9E%E5%9C%A8AC%E4%B8%8A%2CBE%E4%BA%A4CD%E4%BA%8E%E7%82%B9G%2CEF%E2%8A%A5BE%E4%BA%A4AB%E4%BA%8E%E7%82%B9F%2CCE%3DkEA...%E5%A6%82%E5%9B%BE%2C%E5%9C%A8%E2%96%B3ABC%E4%B8%AD%2CAC%3DBC%2C%E2%88%A0ACB%3D90%C2%B0%2CCD%E2%8A%A5AB%2C%E5%9E%82%E8%B6%B3%E4%B8%BAD%2C%E7%82%B9E%E5%9C%A8AC%E4%B8%8A%2CBE%E4%BA%A4CD%E4%BA%8E%E7%82%B9G%2CEF%E2%8A%A5BE%E4%BA%A4AB%E4%BA%8E%E7%82%B9F%2CCE%3DkEA%E2%91%A0%E8%8B%A5AC%3DBC%2C%E6%8E%A2%E7%A9%B6%E7%BA%BF%E6%AE%B5E)
如图,在△ABC中,AC=BC,∠ACB=90°,CD⊥AB,垂足为D,点E在AC上,BE交CD于点G,EF⊥BE交AB于点F,CE=kEA...如图,在△ABC中,AC=BC,∠ACB=90°,CD⊥AB,垂足为D,点E在AC上,BE交CD于点G,EF⊥BE交AB于点F,CE=kEA①若AC=BC,探究线段E
如图,在△ABC中,AC=BC,∠ACB=90°,CD⊥AB,垂足为D,点E在AC上,BE交CD于点G,EF⊥BE交AB于点F,CE=kEA...
如图,在△ABC中,AC=BC,∠ACB=90°,CD⊥AB,垂足为D,点E在AC上,BE交CD于点G,EF⊥BE交AB于点F,CE=kEA
①若AC=BC,探究线段EF与EG的数量关系,并证明.
②若AC=mBC,探究线段EF与EG的数量关系,并证明.
如图,在△ABC中,AC=BC,∠ACB=90°,CD⊥AB,垂足为D,点E在AC上,BE交CD于点G,EF⊥BE交AB于点F,CE=kEA...如图,在△ABC中,AC=BC,∠ACB=90°,CD⊥AB,垂足为D,点E在AC上,BE交CD于点G,EF⊥BE交AB于点F,CE=kEA①若AC=BC,探究线段E
(1)EG=k*EF.
证明:∵AC=BC,∠ACB=90°,CD⊥AB.
∴∠A=∠ACD=45°.
作EM垂直EC,交CD于M,则∠EMC=45°,⊿CEM为等腰直角三角形,EM=CE.
∵∠AEM=∠FEG=90°.
∴∠AEF=∠MEG;又∠A=∠EMG=45°.
∴⊿AEF∽⊿MEG,EF/EG=AE/EM=AE/CE=AE/(k·AE)=1/k,故EG=k·EF
(2)EG=km·EF.
证明:作EM垂直AC,交CD于M.
∵∠CEM=∠ACB=90°;∠ECM=∠CBA(均为∠BCM的余角).
∴⊿CEM∽⊿BCA,CE/EM=BC/CA=BC/(m·BC)=1/m,EM=m·CE.
∵∠AEM=∠FEG=90°.
∴∠AEF=∠MEG;又∠A=∠EMG(均为∠ACD的余角).
故⊿AEF∽⊿MEG,EF/EG=AE/EM=AE/(m·CE)=AE/(m·k·AE)=1/(mk).
所以,EG=mk·EF.
如图(1),在直角△ABC中, ∠ACB=90 ,CD⊥AB,垂足为D,点E在AC上,BE∴∠A=∠EDG,∵EF⊥BE,∴∠AEF+∠FED=∠FED+∠DEG=90°,∴∠AEF