题目内容
如图,AD是△ABC外角∠EAC的平分线,交BC的延长线于点D.延长DA交△ABC的外接圆于点F.
(1)求证:FB=FC;
(2)若FA=2
,AD=4
,求FB的长.
(1)求证:FB=FC;
(2)若FA=2
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3 |
(1)证明:∵A、C、B、F四点共圆
∴∠FBC=∠DAC
又∵AD平分∠EAC
∴∠EAD=∠DAC
又∵∠FCB=∠FAB(同弧所对的圆周角相等),∠FAB=∠EAD
∴∠FBC=∠FCB
∴FB=FC;
(2)∵∠BAC=∠BFC,∠FAB=∠FCB=∠FBC
∴∠FCD=∠BFC+∠FBC=∠BAC+∠FAB=∠FAC
∵∠AFC=∠CFD,
∴△FAC∽△FCD
∴FA:FC=FC:FD
∴FB2=FC2=FA•FD=2
×6
=36,
∴FB=6.
∴∠FBC=∠DAC
又∵AD平分∠EAC
∴∠EAD=∠DAC
又∵∠FCB=∠FAB(同弧所对的圆周角相等),∠FAB=∠EAD
∴∠FBC=∠FCB
∴FB=FC;
(2)∵∠BAC=∠BFC,∠FAB=∠FCB=∠FBC
∴∠FCD=∠BFC+∠FBC=∠BAC+∠FAB=∠FAC
∵∠AFC=∠CFD,
∴△FAC∽△FCD
∴FA:FC=FC:FD
∴FB2=FC2=FA•FD=2
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3 |
∴FB=6.
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