Because AB=AC and AG=AF (known)
So Rt△ABG≌Rt△ACF(HL)
So ∠BAG=∠CAF
So ∠BAG-∠FAG=∠CAF-∠FAG
That is ∠BAF=∠CAG
At △AEF and △ADG,
Because ∠BAF=∠CAG, AF=AG, ∠ AFE = ∠ AGD = 90.
So △ AEF △ ADG (ASA)
So AE=AD