如图,AD为三角形ABC的角平分线,AD的垂直平分线交BC的延长线于点E,交AB于点F.求证AB方比AC方=BE比CE
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![如图,AD为三角形ABC的角平分线,AD的垂直平分线交BC的延长线于点E,交AB于点F.求证AB方比AC方=BE比CE](/uploads/image/z/10156263-15-3.jpg?t=%E5%A6%82%E5%9B%BE%2CAD%E4%B8%BA%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E7%9A%84%E8%A7%92%E5%B9%B3%E5%88%86%E7%BA%BF%2CAD%E7%9A%84%E5%9E%82%E7%9B%B4%E5%B9%B3%E5%88%86%E7%BA%BF%E4%BA%A4BC%E7%9A%84%E5%BB%B6%E9%95%BF%E7%BA%BF%E4%BA%8E%E7%82%B9E%2C%E4%BA%A4AB%E4%BA%8E%E7%82%B9F.%E6%B1%82%E8%AF%81AB%E6%96%B9%E6%AF%94AC%E6%96%B9%3DBE%E6%AF%94CE)
如图,AD为三角形ABC的角平分线,AD的垂直平分线交BC的延长线于点E,交AB于点F.求证AB方比AC方=BE比CE
如图,AD为三角形ABC的角平分线,AD的垂直平分线交BC的延长线于点E,交AB于点F.求证AB方比AC方=BE比CE
如图,AD为三角形ABC的角平分线,AD的垂直平分线交BC的延长线于点E,交AB于点F.求证AB方比AC方=BE比CE
郭敦顒回答:
未见图,依题意自绘了.
AD是△ABC角的平分线,AD的垂直平分线交BC的延长线于E,交AB于F,
求证:AB²:AC²=BE:CE
AC与EF交于G,AF=AG=DF=DG,四边形AFDG为菱形,
DF∥AC,DG∥AB,
∴DG/BF=DE/BE (1)
作CK⊥EF于K,则Rt⊿CKG∽Rt⊿AOG,
CG/AG=CK/AO,
∵Rt⊿ECK∽Rt⊿EDO,
∴CK/DO=CE/DE,∵AO=DO,
∴CG/AG=CE/DE (2)
(1)•(2)得
(CG/AG)•(DG/BF)=(CE/DE)•(DE/BE)=CE/BE
(CG•DG )/(AG•BF)=CE/BE,
(CG•AF )/(AF•BF)=CE/BE,
∴CE/BE=CG/BF (3)
∵△ABC∽△FBD∽△GDC,
∴BF/AB=DF/AC,
DG/AB=CG/AC,
∴BF• DG /AB²=DF•CG /AC²,
∴BF• AF /AB²=AF•CG /AC²,
∴AC²/AB²=(AF•CG )/ (BF• AF)=CG/BF ,
∴AC²/AB²= CG/BF (4)
由(3),(4)得,
AC²/AB²= CE/BE
∴AB² :AC²= BE:CE.
A
F
O G
K
B D C E
题有毛病 亲~~ 看不懂 再说没图