∠DAE,又∠EAC=∠EBC,∴∠DAC=∠ABC,∵AD∥BC,∴∠DAC=∠ACB,∴∠ABC=∠ACB,∴AB=AC2作AF⊥CD于F,∵四边形ABCE是圆内接四边形,∴∠ABC=∠AEF,又∠ABC=∠ACB,∴∠AEF=∠ACB,
∠AHE=∠AFE,又∠AEB=∠ACB,∴∠AEH=∠AEF,在△AEH和△AEF中,∠AEH=∠AEF,∴△AEH≌△AEF,
AE=AE,
∠ABH=∠ACF,∴EH=EF,∴CE+EH=CF,在△ABH和△ACF中,∠AHB=∠AFC,∴△ABH≌△ACF,∴BH
AB=AC,=CF=CE+EH
5.2015来宾已知⊙O是以AB为直径的△ABC的外接圆,OD∥BC交⊙O于点D,交AC于点E,连接AD,BD,BD交AC于点F
f1求证:BD平分∠ABC;2延长AC到点P,使PF=PB,求证:PB是⊙O的切线;3如果AB=10,cos∠ABC=35,求AD
解:1∵OD∥BC,∴∠OBD=∠CBD,∵OB=OD,∴∠OBD=∠OBD,∴∠CBD=∠OBD,∴BD平分∠ABC2∵⊙O是以AB为直径的△ABC的外接圆,∴∠ACB=90°,∴∠CFB+∠CBF=90°∵PF=PB,∴∠PBF=∠CFB,由1知∠OBD=∠CBF,∴∠PBF+∠OBD=90°,∴∠OBP=90°,∴PB是⊙O的切线3∵在Rt△ABC中,∠ACB=90°,AB=10,∴cos∠ABC=BACB=B1C0=35,∴BC=6,AC=AB2-BC2=8∵OD∥BC,∴△AOE∽△ABC,∠AED=∠OEC=180°-∠ACB=90°,∴AAEC=OBEC=AABO,A8E=O6E=150,∴AE=4,OE=3,∴DE=OD-OE=5-3=2,∴AD=AE2+DE2=42+22=25
6.2015南宁如图,AB是⊙O的直径,C,G是⊙O上两点,且AC=CG,过点C的直线CD⊥BG于点D,交BA的延长线于点E,连接BC,交OD于点F1求证:CD是⊙O的切线;2若OFFD=23,求∠E的度数;3连接AD,在2的条件下,若CD=3,求AD的长.
解:1如图1,连接OC,AC,CG,∵AC=CG,∴∠ABC=∠CBG,∵OC=OB,∴∠OCB=∠OBC,∴∠OCB=∠CBG,∴OC∥BG,∵CD⊥BG,∴OC⊥CD,∴CD是⊙O的切线2∵
fOC∥BD,∴△OCF∽△BDF,△EOC∽△EBD,∴BODC=ODFF=23,∴OBCD=OBEE=23,∵OA=OB,∴AE=OA=OB,∴OC=12OE,∵∠ECO=90°,∴∠E=30°3如图2,过A作AH⊥DE于H,∵∠E=30°,∴∠EBD=60°,∴∠CBD=12∠EBD=30°,∵CD=3,∴BD=3,DE=33,BE=6,∴AE=31BE=2,∴AH=1,∴EH=3,∴DH=23,在Rt△DAH中,AD=AH2+DH2=
12+(23)2=13
fr