已知:如图,角BAC=90度,AD垂直BC于点D,BG平分角ABC交AD于E,交AC于G,EF//BC交AC于F.求证：AE=CF
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已知:如图,角BAC=90度,AD垂直BC于点D,BG平分角ABC交AD于E,交AC于G,EF//BC交AC于F.求证：AE=CF
已知:如图,角BAC=90度,AD垂直BC于点D,BG平分角ABC交AD于E,交AC于G,EF//BC交AC于F.求证：AE=CF
已知:如图,角BAC=90度,AD垂直BC于点D,BG平分角ABC交AD于E,交AC于G,EF//BC交AC于F.求证：AE=CF
证明：∵∠ABG=∠CBG∴AG/GC=AB/BC
∵AD⊥BC∴∠ADB=90°∴∠BED+∠CBG=90°∵∠BAC=90°∴∠AGE+ABG=90°∴∠AGE=∠BED∵∠BED=∠AEG∴∠AEG=∠AGE∴AE=AG
∵EF∥BC∴∠AEF=∠ADC=90°,∠AFE=∠C∴△AEF∽△BAC∴AB/BC=EA/AF=AG/AF
∴AB/BC=AG/GC=AG/AF∴GC=AF∴AG=CF
∴AE=CF
设AB=1,则AD=sinB,ED=cosB,ED=BD*tan(B/2)=cosB*tan(B/2),
AE=AD-ED
=sinB- cosB*tan(B/2)=2sin(B/2)*cos(B/2)-(cos(B/2)^2-sin(B/2)^2)*tan(B/2)
=2sin(B/2)*cos(B/2)/(sin(B/2)^2+cos(B/2)^2)-(cos(B/2)...
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设AB=1,则AD=sinB,ED=cosB,ED=BD*tan(B/2)=cosB*tan(B/2),
AE=AD-ED
=sinB- cosB*tan(B/2)=2sin(B/2)*cos(B/2)-(cos(B/2)^2-sin(B/2)^2)*tan(B/2)
=2sin(B/2)*cos(B/2)/(sin(B/2)^2+cos(B/2)^2)-(cos(B/2)^2-sin(B/2)^2)/(sin(B/2)^2+cos(B/2)^2)* tan(B/2)
=2 tan(B/2)/(1+ tan(B/2)^2)-(1- tan(B/2)^2)/ (1+ tan(B/2)^2)*tan(B/2)
=[2/(1+ tan(B/2)^2)-(1- tan(B/2)^2)/ (1+ tan(B/2)^2)]* tan(B/2)
=[2-(1- tan(B/2)^2)]/ (1+ tan(B/2)^2)* tan(B/2)
=[(1+tan(B/2)^2)]/ (1+ tan(B/2)^2)* tan(B/2)
= tan(B/2)
CF=ED/cosB= tan(B/2)
所以，AE=CF
收起
如图作2条垂线，等量关系在图中标注了，过程略