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.
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(1)即BE平分∠ABC;(2)EF=
.
解析:
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=
.
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