(20 12? Suzhou) As shown in the figure, in the trapezoidal ABCD, it is known that AD∨BC and AB=CD, and the line segment CB is extended to E, so that BE=AD is used to connect AE and AC. (1) for verific

(20 12? Suzhou) As shown in the figure, in the trapezoidal ABCD, it is known that AD∨BC and AB=CD, and the line segment CB is extended to E, so that BE=AD is used to connect AE and AC. (1) for verification. (1) proves that in trapezoidal ABCD, ∫AD∨BC, AB=CD,

∴∠ABE=∠BAD,∠BAD=∠CDA，

∴∠ABE=∠CDA

In △ABE and △CDA, AB = CD ∠ Abe = ∠ Be = dacda,

∴△ABE≌△CDA.

(2) Solution: obtained from (1): ∠AEB=∠CAD, AE=AC,

∴∠AEB=∠ACE，

∫∠DAC = 40，

∴∠AEB=∠ACE=40，

∴∠eac= 180-40-40 = 100。