已知点d在三角形abc边bc上,且与b,d不重合

1个回答

  • (1)∵DE∥AC,DF∥AB,

    ∴△BDE∽△BCA∽△DCF,

    记S△BDE=S1,

    S△DCF=S2,

    ∵SAEFD= 2/5S,

    ∴S1+S2=S- 2/5S= 3/5S.①

    √S1/√S= BD/BC, √S2/√S= CD/BC,

    于是 √S1/S+ √S2/S= ﹙BD+CD﹚/BC=1,即 √S1+√ S2= √S,

    两边平方得S=S1+S2+2 √﹙S1S2﹚,

    故2 √﹙S1S2﹚=SAEFD= 2/5S,即S1S2= 1/25S2.②

    由①、②解得S1= ﹙3±√5﹚/10S,即 S1/S=﹙ 3±√5﹚/10.

    而 S1/S= (BD/BC)²,即 ﹙3±√5﹚/10= (BD/5)2,解得BD= ﹙5±√5﹚/2.

    (2)由G是△ABC的重心,DF过点G,且DF∥AB,可得 CD/CB= 2/3,则DF= 2/3AB.

    由DE∥AC, CD/CB= 23,得DE= 1/3AC,

    ∵AC= ∨2AB,∴ AC/AB=√ 2, DF/ED= 2AB/√﹙2AB﹚=√ 2,

    得 DF/DE= AC/AB,即 DF/AC= DE/AB,

    又∠EDF=∠A,故△DEF∽△ABC,

    得 EF/BC= DE/AB,所以EF= 5√2/3