如图,点D在△ABC的边BC上,过点D作AC的平行线DE交AB于E,作AB的平行线AE交AC于FBC=5(1)设△ABC的面积为S,若四边形AEDF的面积为2/5S,求BD的长(2)若AB=根号2AC,AM为△ABC的BC边上的中线,G为重心,当DE经
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![如图,点D在△ABC的边BC上,过点D作AC的平行线DE交AB于E,作AB的平行线AE交AC于FBC=5(1)设△ABC的面积为S,若四边形AEDF的面积为2/5S,求BD的长(2)若AB=根号2AC,AM为△ABC的BC边上的中线,G为重心,当DE经](/uploads/image/z/6076347-51-7.jpg?t=%E5%A6%82%E5%9B%BE%2C%E7%82%B9D%E5%9C%A8%E2%96%B3ABC%E7%9A%84%E8%BE%B9BC%E4%B8%8A%2C%E8%BF%87%E7%82%B9D%E4%BD%9CAC%E7%9A%84%E5%B9%B3%E8%A1%8C%E7%BA%BFDE%E4%BA%A4AB%E4%BA%8EE%2C%E4%BD%9CAB%E7%9A%84%E5%B9%B3%E8%A1%8C%E7%BA%BFAE%E4%BA%A4AC%E4%BA%8EFBC%3D5%EF%BC%881%EF%BC%89%E8%AE%BE%E2%96%B3ABC%E7%9A%84%E9%9D%A2%E7%A7%AF%E4%B8%BAS%2C%E8%8B%A5%E5%9B%9B%E8%BE%B9%E5%BD%A2AEDF%E7%9A%84%E9%9D%A2%E7%A7%AF%E4%B8%BA2%2F5S%2C%E6%B1%82BD%E7%9A%84%E9%95%BF%EF%BC%882%EF%BC%89%E8%8B%A5AB%3D%E6%A0%B9%E5%8F%B72AC%2CAM%E4%B8%BA%E2%96%B3ABC%E7%9A%84BC%E8%BE%B9%E4%B8%8A%E7%9A%84%E4%B8%AD%E7%BA%BF%2CG%E4%B8%BA%E9%87%8D%E5%BF%83%2C%E5%BD%93DE%E7%BB%8F)
如图,点D在△ABC的边BC上,过点D作AC的平行线DE交AB于E,作AB的平行线AE交AC于FBC=5(1)设△ABC的面积为S,若四边形AEDF的面积为2/5S,求BD的长(2)若AB=根号2AC,AM为△ABC的BC边上的中线,G为重心,当DE经
如图,点D在△ABC的边BC上,过点D作AC的平行线DE交AB于E,作AB的平行线AE交AC于F
BC=5
(1)设△ABC的面积为S,若四边形AEDF的面积为2/5S,求BD的长
(2)若AB=根号2AC,AM为△ABC的BC边上的中线,G为重心,当DE经过G时△ABC和△ABC相似吗?,求出此时EF的长,若不相似,说明理由.
如图,点D在△ABC的边BC上,过点D作AC的平行线DE交AB于E,作AB的平行线AE交AC于FBC=5(1)设△ABC的面积为S,若四边形AEDF的面积为2/5S,求BD的长(2)若AB=根号2AC,AM为△ABC的BC边上的中线,G为重心,当DE经
(1)∵DE∥AC,DF∥AB,
∴△BDE∽△BCA∽△DCF,
记S△BDE=S1,S△DCF=S2,
∵SAEFD=25S,
∴S1+S2=S-25S=35S.①
S1S=BDBC,S2S=CDBC,
于是S1S+S2S=BD+CDBC=1,即S1+S2=S,
两边平方得S=S1+S2+2S1S2,
故2S1S2=SAEFD=25S,即S1S2=125S2.②
由①、②解得S1=3±
510S,即S1S=3±
510.
而S1S=(
BDBC)2,即3±
510=(
BD5)2,解得BD=30±10
52=(5±
5)22=5±
52.
(2)由G是△ABC的重心,DF过点G,且DF∥AB,可得CDCB=23,则DF=23AB.
由DE∥AC,CDCB=23,得DE=13AC,
∵AC=2AB,∴ACAB=2,DFED=2AB2AB=2,
得DFDE=ACAB,即DFAC=DEAB,
又∠EDF=∠A,故△DEF∽△ABC,
得EFBC=DEAB,所以EF=5
23.