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Changning district 20 16 bimodule mathematics
(1) proof: ∫AD∨BE, BE∨AC,

∴ACED is a parallelogram,

∴AC=DE,

∵ isosceles trapezoid ABCD,

∴AC=BD,

∴BD=DE?

∵AC⊥BD,

∴∠BOC=90,

∫AC∑DE,

∴∠BOC=∠BDE=90,

∴△BDE is an isosceles right triangle.

(2) solution: ∫AD∨BC,

∴OAOC=ODOB=ADBC,

∴ACOC=BDOB

∵ isosceles trapezoid ABCD,

∴AC=BD,

∴OC=OB,OA=OD,

∫DE∑AC,

∴∠CDE=∠DCO,

∴sin∠CDE=sin∠DCO=55,

At Rt△DCO, let OD=k and DC=5k? (k > 0), then OC=DC2? OD2=2k,

Parallelogram ACDE,

∴AD=CE,

∴ODOB=ODOC= 12,

∴ADBC= 12,

∴ADBE= 13.