6. 如图,在△ABC中,∠C=90°,AD是∠BAC的平分线,DE⊥AB于点E,点F在AC上,BD=DF.求证:
(1)CF=EB;
(2)AB=AF+2EB.
(1)CF=EB;
(2)AB=AF+2EB.
答案:
(1)证明:
∵∠C=90°,AD是∠BAC的平分线,DE⊥AB,
∴DC=DE,∠C=∠DEB=90°,
在Rt△DCF和Rt△DEB中,
∵DF=BD,DC=DE,
∴Rt△DCF≌Rt△DEB(HL),
∴CF=EB;
(2)证明:在Rt△ADC和Rt△ADE中,
∵AD=AD,DC=DE,
∴Rt△ADC≌Rt△ADE(HL),
∴AC=AE,
∵AB=AE+EB,AC=AF+CF,CF=EB,
∴AB=AF+CF+EB=AF+EB+EB=AF+2EB.
∵∠C=90°,AD是∠BAC的平分线,DE⊥AB,
∴DC=DE,∠C=∠DEB=90°,
在Rt△DCF和Rt△DEB中,
∵DF=BD,DC=DE,
∴Rt△DCF≌Rt△DEB(HL),
∴CF=EB;
(2)证明:在Rt△ADC和Rt△ADE中,
∵AD=AD,DC=DE,
∴Rt△ADC≌Rt△ADE(HL),
∴AC=AE,
∵AB=AE+EB,AC=AF+CF,CF=EB,
∴AB=AF+CF+EB=AF+EB+EB=AF+2EB.
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