正方形ABCD中,E为CD上一点,AE交BD于F,FG⊥AE于F,交BC于G,PF⊥PD于G,求证:PF=1/2 BD

2个回答

  • 自己画辅助线

    1)过F做FM//BE,交AE于M

    BE=2/3*BE=2/3*AD

    所以 PF/PD=FM/AD=2/3*FM/BE=2/3*AF/AB=2/3*1/(1+n)=2/(3n+3)

    所以AP/PM=PD/PF=(3n+3)/2

    所以AP/AM=(3n+3)/(3n+5)

    又AM/AE=AF/AB=1/(n+1)

    所以AP/AE=AP/AM*AM/AE=3/(3n+5)

    所以AP/PE=3/(3n+2)

    所以:

    当n=1时,PF/PD=1/3,AP/PE=3/5

    2)n=2时,AP/PE=3/(3n+2)=3/8

    所以8AP=3PE

    3)不妨设正方形边长为1

    勾股定理:DF=√(1+1/(n+1)^2)

    PF/PD=2/(3n+3)

    PD/DF=(3n+3)/(3n+5)

    PD=DF*(3n+3)/(3n+5)

    PD*DF=(3n+3)/(3n+5)*DF^2=3(n^2+2n+2)/(n+1)(3n+5)

    AE⊥DF 是用摄影定理有PD*DF=AD^2=1

    所以3(n^2+2n+2)/(n+1)(3n+5)=1

    解得n=1/2