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