如图,在Rt△ACB中,∠ACB=90°,点D在边AB上,DE平分∠CDB交边BC于E,EM是线段BD的垂直平分线.

1个回答

  • (1)证明:∵EM是线段BD的垂直平分线,

    ∴ED=EB,

    ∴∠EDB=∠B,

    ∵DE平分∠CDB,

    ∴∠CDE=∠EDB,

    ∴∠CDE=∠B,

    ∵∠DCE=∠BCD,

    ∴△CDE ∽ △CBD,

    CD

    BC =

    DE

    BD ,

    ∵ED=EB,

    CD

    BC =

    BE

    BD ;

    (2)∵∠ACB=90°,AB=10,cosB=

    4

    5 ,

    ∴AC=6,BC=8,

    ∵EM是线段BD的垂直平分线,

    ∴DM=BM,

    CD

    BC =

    BE

    BD =

    BE

    2BM ,

    CD

    8 =

    BE

    2BM ,

    即CD=

    4BE

    BM ,

    ∵cosB=

    BM

    BE =

    4

    5 ,

    ∴CD=4×

    5

    4 =5.