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Obviously I prefer @Michael Rozenberg's solution, as its straightforward to adapt for E, F, G, H not middle-points (clearly if AH=0 then the inner square is the outer square and the answer is 1; the posed question is AH=1/2 with answer 1/5; the other limit AH=AD=1 then clearly the answer is 0, but what of the intermediate values?).

So, to visually answer the question I'd go with a variation on the solution by @DeltaScuti_Fomalhautb. My large pink triangles are clearly each 1/4, so if subtracting them each from the main square (1-4* 1/4 =0), the leftover inner square is exactly what I've double-subtracted (the four darker shaded overlaps, one in each corner of the original square; ignore the extra stripey overlay).

Now to show that five of these double-counted triangles together make up my large pink triangles is obvious: The top-left pink triangle consists of the 'stripey-shaded' medium-sized triangle plus one of the overlappers, and the 'stripey-shaded' one is exactly four times the little overlapper it contains at it's right-hand point (it's congruent with the little one, and its hypothenuse is exactly twice the little one's hypothenuse, so its surface is four times the little one's). So in summary this proves the little square is four times the little overlap-triangle, each of which is 1/5 of the pink triangle which was 1/4 of the original square; so the little square is 4* 1/5 * 1/4 = 1/5.

Ignore the blue dotted lines, they'd give you another way to see it's five of those little triangles, but it's slightly more work to argue --- e.g., at this point I haven't established that the inner square's side is equal to the overlapper's base, and the dotted lines thus define shapes that seem congruent with the overlappers but might not be (like in that trick where you divide a rectangle into triangles but 1/16 or so gets lost as two lines aren't exactly parallel though drawing in fat pencil they seem; sorry can't find a link now).