题目内容
如图,⊙O是等腰三角形ABC的外接圆,AB=AC,延长BC到点D,使CD=AC,连接AD交⊙O于点E,连接BE与AC交于点F.
⑴判断BE是否平分∠ABC,并说明理由;
⑵若AE=6,BE=8,求EF的长.
⑴判断BE是否平分∠ABC,并说明理由;
⑵若AE=6,BE=8,求EF的长.

(1)即BE平分∠ABC;(2)EF=
.

⑴BE平分∠ABC.
∵CD=AC,∴∠D="∠CAD."
∵AB=AC,∴∠ABC=∠ACB
∵∠EBC=∠CAD,∴∠EBC="∠D=∠CAD. "
∵∠ABC=∠ABE+∠EBC,∠ACB=∠D+∠CAD,
∴∠ABE=∠EBC,即BE平分∠ABC.
⑵由⑴知∠CAD="∠EBC" =∠ABE.
∵∠AEF=∠AEB,∴△AEF∽△BEA.
∴
,∵AE=6, BE=8.
∴EF=
.
∵CD=AC,∴∠D="∠CAD."
∵AB=AC,∴∠ABC=∠ACB
∵∠EBC=∠CAD,∴∠EBC="∠D=∠CAD. "
∵∠ABC=∠ABE+∠EBC,∠ACB=∠D+∠CAD,
∴∠ABE=∠EBC,即BE平分∠ABC.
⑵由⑴知∠CAD="∠EBC" =∠ABE.
∵∠AEF=∠AEB,∴△AEF∽△BEA.
∴

∴EF=


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