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Senior three blackboard mathematics
(1) If △BEC is an isosceles triangle, then BE=CE is solvable.

If the two extracted ones are ① ②, we can get △ Abe△ DCE, be = ec.

If ① ③, AE=DE, AB=CD, △ Abe △ DCE cannot be obtained, ∴ This situation is not true;

If ① ④, we can get △ Abe△ DCE, be = ec.

If ② ③, you can also get △ Abe△ DCE, be = ec.

If ② ④, the three angles are equal, but the side lengths are not necessarily equal, it is not true.

If ③ ④, be = EC can also be obtained.

So, the answer is: ① ③; ②④.

(2) Using ① and ② as conditions, it can be judged that △BCE is an isosceles triangle.

∫AB = DC∠ABE =∠DCE,

∫∠AEB =∠DEC again.

∴△ABE≌△DCE(AAS),

∴BE=EC, that is, △BCE is an isosceles triangle.