1.已知梯形ABCD,AD平行于BC,AC、BD相交于O点,G是BD的中点,过G作EF平行于AC交AB于E,交BC于F,若AC=10,AD=3分之1BC.求证:EF的长http://hiphotos.baidu.com/%C0%BC%C9%AB%CA%A5%D3%F2/abpic/item/701f06512d8033958d54303b.jpg2.三角
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/26 01:12:23
![1.已知梯形ABCD,AD平行于BC,AC、BD相交于O点,G是BD的中点,过G作EF平行于AC交AB于E,交BC于F,若AC=10,AD=3分之1BC.求证:EF的长http://hiphotos.baidu.com/%C0%BC%C9%AB%CA%A5%D3%F2/abpic/item/701f06512d8033958d54303b.jpg2.三角](/uploads/image/z/12511997-53-7.jpg?t=1.%E5%B7%B2%E7%9F%A5%E6%A2%AF%E5%BD%A2ABCD%2CAD%E5%B9%B3%E8%A1%8C%E4%BA%8EBC%2CAC%E3%80%81BD%E7%9B%B8%E4%BA%A4%E4%BA%8EO%E7%82%B9%2CG%E6%98%AFBD%E7%9A%84%E4%B8%AD%E7%82%B9%2C%E8%BF%87G%E4%BD%9CEF%E5%B9%B3%E8%A1%8C%E4%BA%8EAC%E4%BA%A4AB%E4%BA%8EE%2C%E4%BA%A4BC%E4%BA%8EF%2C%E8%8B%A5AC%3D10%2CAD%3D3%E5%88%86%E4%B9%8B1BC.%E6%B1%82%E8%AF%81%EF%BC%9AEF%E7%9A%84%E9%95%BFhttp%3A%2F%2Fhiphotos.baidu.com%2F%25C0%25BC%25C9%25AB%25CA%25A5%25D3%25F2%2Fabpic%2Fitem%2F701f06512d8033958d54303b.jpg2.%E4%B8%89%E8%A7%92)
1.已知梯形ABCD,AD平行于BC,AC、BD相交于O点,G是BD的中点,过G作EF平行于AC交AB于E,交BC于F,若AC=10,AD=3分之1BC.求证:EF的长http://hiphotos.baidu.com/%C0%BC%C9%AB%CA%A5%D3%F2/abpic/item/701f06512d8033958d54303b.jpg2.三角
1.已知梯形ABCD,AD平行于BC,AC、BD相交于O点,G是BD的中点,过G作EF平行于AC交AB于E,交BC于F,若AC=10,AD=3分之1BC.
求证:EF的长
http://hiphotos.baidu.com/%C0%BC%C9%AB%CA%A5%D3%F2/abpic/item/701f06512d8033958d54303b.jpg
2.三角形ABC中,角ABC=2角C,D在AC上,联结BD,如果AB的平方=AD*AC.
求证:BD是角ABC的平分线
http://hiphotos.baidu.com/%C0%BC%C9%AB%CA%A5%D3%F2/abpic/item/a01049fe7f7e3b2b5d600838.jpg
1.已知梯形ABCD,AD平行于BC,AC、BD相交于O点,G是BD的中点,过G作EF平行于AC交AB于E,交BC于F,若AC=10,AD=3分之1BC.求证:EF的长http://hiphotos.baidu.com/%C0%BC%C9%AB%CA%A5%D3%F2/abpic/item/701f06512d8033958d54303b.jpg2.三角
1.在梯形ABCD中
∵AD‖BC,∴△AOD∽△COB,
∴DO:BO=AD:CB=1/3
∴BO=3BD/4
又∵G为BD中点
∴BG=BD/2
从而BG:BO=2:3
在△ABO和△EBG中,EF‖AC,∠ABO为公共角
∴△ABO∽△EBG
∴BE:BA=BG:BO=2:3
同理△ABC∽△EBF
∴EF:AC=BE:BA=2:3
∵AC=10(已知)
∴EF=2AC/3=20/3
2.证明:∵AB^2=AD*AC(已知)
∴AD:AB=AB:AC
又∵在△BAD和△CAB中,∠A是公共角
∴△BAD∽△CAB
∴∠ABD=∠BCA
又∵∠ABC=2∠BCA(已知)
∴∠ABC=2∠BCA=2∠ABD
故BD为∠ABC角平分线