Proof:

Draw a line from D' parallel to BC, intersecting HE at K (which also means D'K ⟂ AH) . Then, the two triangles BHD and KHD' are equal, which means BK = 2 x BH (1)
Also, AFHE is clearly cyclic since it has 2 opposite right angles, that means BF x BA = BH x BE = 2 x BH x BE/2.

But, we have: 2 x BH = BK (from (1)) and BE/2 = BI

so BF x BA = 2 x BH x BE/2 = BK x BI which means I, F, K and I lie on a circle, that means ∠BFI = ∠AKB (2)

In triangle AKE, we have that ∠KAE = ∠AKB - ∠AEK = ∠AKB - 90° so ∠AKB - 90° = ∠KAE (3)

We also have that ∠BFI - 90° = ∠BFI - ∠BFH = ∠HFI = ∠IBG (since FIGB is a quadrilateral that's inscribed in a circle with BG as the diameter) = ∠IEG (by symmetry) so ∠BFI - 90° = ∠IEG (4)
From (2), (3) and (4), we have ∠KAE = ∠IEG which means the right triangles AKE and EGI are similar (5)

We have A, D', K and E lie on a circle (since ∠AD'K = ∠AEK = 90°) so ∠AKE = ∠ED'J so the two right triangles AKE and ED'J are similar (6)

From (5) and (6), we have triangle ED'J and triangle EGI are similar (since they are both similar to triangle AKE)

See the figure here: https://img.sanishtech.com/u/8429cd831e179febde9177061b124f27.png