已知三角形ABC,分别以AB、AC为边向形外作等边△ABF、△ACE,再以AF、AE为边作平行四边形AEDF,求证三角形

3个回答

  • 在平行四边形AEDF中,有:AE = FD ;

    所以,AC = AE = FD .

    若∠ABC > 60°,则有:∠ABC = 60°+∠ABD = ∠FBD ;

    若∠ABC < 60°,则有:∠ABC = 60°-∠ABD = ∠FBD ;

    所以,∠ABC = ∠FBD .

    在△ABC和△FBD中,AB = FB ,∠ABC = ∠FBD ,AC = FD ,

    所以,△ABC ≌ △FBD ;

    可得:BC = BD .

    同理可得:BC = CD ,

    则有:BC = BD = CD ,

    即:△BCD是等边三角形.