(1) Because ab//cd of parallelogram abcd, ∠abc=∠ecf, ∠aeb=∠cef (diagonally equal).

So ⊿abe≡⊿fce (I don't know if ⊿abe≡⊿fce is similar or not, just add the following conditions of 2 to the congruence. )

(2) Because ⊿abe≡⊿fce and be=ec, cf=ab, cf//ab. The quadrilateral abcf is a parallelogram.

∠ AEB =180-∠ AEC =180-2 ∠ abcf, and ∠ BAE =180-∠ Abe-∠.

Hope to adopt!