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Jing 'an 20 15 Mathematics Joint Examination Module 2
Proof: (1) Connect the communication,

∫ In trapezoidal ABCD, AD∨BC,

∴∠EAD=∠B,

In △DEA and △ABC,

AE=BC∠EAD=∠BAD=AB,

∴△DEA≌△ABC(SAS),

∫∠AED =α,

∴∠BCA=∠AED=α,

AD = CD,

∴∠DCA=∠DAC=∠ACB=α,

∴∠bcd=∠dca+∠acb=2α;

(2) Ed ∠BEC,

∴∠AEC=2∠AED=2α.

∫ ABCD, AD∨BC, AB=CD in the trapezoid,

∴∠EAD=∠B=∠BCD=2α=∠AEC,

∴CE=BC=AE,

∴∠ECA=∠EAC=∠EAD+∠DAC=3α,

∴∠ECB=∠ECA+∠ACB=4α.

∵∠B+∠BEC+∠BCE= 180,

∴2α+2α+4α= 180 ,

∴∠ECB=4α=90。

∴△EBC is an isosceles right triangle.