Extend AE to e', so that EE'=AC, even BE', then AG=GE', HG is the center line of △ABE', and HG=BE'/2.

Extend BD to d' so that DD'=BC, even AD', then BF=FD', HF is the center line of △BAD', and HF=AD'/2.

Because AD=AC, ∠ADC=∠ACD.

Because BC=BE, ∠BEC=∠BCE.

And diagonal ∠ACD=∠BCE.

Therefore, ADC=∠BEC.

Therefore, their complementary angles ∠ADD'=∠E'EB.

DD'=BC=EB

AD=AC=E'E

So △ add '△ e 'eb

So, AD = BE

And HG=BE'/2, HF=AD'/2.

Therefore, HG=HF.