From △ACB is an isosceles right triangle and D is the midpoint of AB, we get

CD= 1/2AB=BD，∠KCD=∠B=45

From ∠ CDB = 90 = ∠ EDF, ∠KDC=∠HDB is obtained.

So △ kcd △ hbd, KC=HB, KD=DH.

Because DG divides ∠EDF equally, we get △ kgd △ hgd, KG=GH.

From ∠ kcg = 90, kg 2 = KC 2+CG 2, so GH 2 = HB 2+CG 2.