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