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.
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(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.