(拓展深化)如图①所示,△ABC内接于⊙O,AB=AC,D是BC边上的一点,E是直线AD和△ABC外接圆的交点.

1个回答

  • 见解析

    证明 (1)如图③,连接BE.

    ∵AB=AC,∴∠ABC=∠ACB.

    ∵∠ACB=∠AEB,

    ∴∠ABC=∠AEB.

    ∴△ABD∽△AEB.

    ∴AB∶AE=AD∶AB,

    即AB 2=AD·AE.

    (2)如图④,连接BE、EC,

    ∵四边形ABCE内接于⊙O,

    ∴∠CED=∠ABC,

    ∵AB=AC,∴∠ABC=∠ACB,

    ∴∠CED=∠ACB,

    ∵∠AEC=180°-∠CED,

    ∠ACD=180°-∠ACB,

    ∴∠AEC=∠ACD,∴△ACE∽△ADC,

    ∴=,∴AB 2=AD·AE.