如图,⊙O是等腰三角形ABC的外接圆,AB=AC,延长BC到点D,使得CD=AC,连接AD交⊙O于点E,连接BE与AC交于点F,求证BE平分∠ABC.

问题描述:

如图,⊙O是等腰三角形ABC的外接圆,AB=AC,延长BC到点D,使得CD=AC,连接AD交⊙O于点E,连接BE与AC交于点F,求证BE平分∠ABC.

证明:∵CD=AC,
∴∠D=∠CAD.
∵AB=AC,
∴∠ABC=∠ACB.
∵∠EBC=∠CAD,
∴∠EBC=∠D.
∵∠ABC=∠ABE+∠EBC,∠ACB=∠D+∠CAD.
∴∠ABE=∠EBC,
即BE平分∠ABC.