△ABC是等边三角形,∠DAE=120°求证BC的平方=DB×CE 急，在线等。

显示全部楼层 · 2011-5-11 00:39:28

∵△ABC是等边三角形∴AB=AC=BC∠ABC=∠ACB=∠BAC=60°∴∠D+∠DAB=∠ABC=60°
（1）∠E+∠CAE=∠ACB=60°
（2）∵∠DAE=120°∴∠BAD+∠CAE=120-60°=60°（3）综合(1)(2)(3)知∠D=∠CAE∠DAB=∠E∴△ADB∽△EAC∴DB/AC=AB/CEAC*AB=BC^2=DB*CE

千问 · 2011-5-11 00:39:28

∠ADB+∠DAB=∠ABC=60°；∠DAB+CAE=∠DAE-∠BAC=120°-60°=60°；=>∠ADB+∠DAB=∠DAB+CAE=60° =>∠ADB=∠CAE；同理=>∠DAE=∠E ∴ △ADB∽△EAC =>AB：CE=DB：AC=>BC：CE=DB：BC =>BC2=DB×CE

千问 · 2011-5-11 00:39:28

∵∠E+∠EAC=60∠DAB+∠EAC=120-60=60∠E=∠DAB同理∠D=∠EAC△ADB∽△EACBD/AC=AB/CEAB*AC=BD*CEBC2=BD*CE