∵ AD//BC, AF⊥BC is at point F.

∴∠ADG=∠CBG ①

∠BAE =∠AFB-∠ABF =∠AFB-∠ABC = 90-75 = 15②

AF⊥AD

So AG is the center line of the hypotenuse DE of the right triangle ADE.

∴AG= 1/2DE=GD=GE=AB

Then ∠ Dag =∠ADG, ∠AGB=∠ABG.

Let ∠ADG=X

Then ∠DAG=∠GDA=X,

∠ AGB =∠ ABG =∠ ABC-∠ CBG =∠ ABC-∠ ADG = 75-x [derived from ① ∠ADG=∠CBG] ③.

And ∠AGB is the outer corner of the triangle ADG.

Therefore, ∠AGB =∠ Dag +∠ADG=X+X=2X ④.

75-x = 2x from ③ ④

∴X=25

So ∠ abg = ∠ ABC-∠ cbg = 75-x = 75-25 = 50 ⑤.

∴ AED =∠ BAE +∠ ABG = 15+50 = 65 from ② ⑤.