Let AD be the center line of BC side in △ABC.

Get BD=CD

Because CE‖AB

It is concluded that ∠ABD=∠DCE, ∠BAD=∠CED (the lines are parallel and the internal dislocation angles are equal).

BD=CD，∠ABD=∠DCE，∠BAD=∠CED

Get △ Abd△ CDE.

So AB=CE

(2)

AD=DE is obtained from △ Abd △ CDE.

So 2AD=AD+DE=AE.

△AC+CE > in△ACE； AE (the sum of two sides of a triangle is greater than the third side)

So AC+CE >；; 2AD

And because AB=CE (proved)

So 2ad