For reference: If ABCD and EFGH are squares, HI = 2 and GN=$\sqrt5$ , calculate EH.
My progress:
If point I were given as the midpoint of $AD$ the exercise would be quite easy $AE=4=AI, EH = \sqrt{(4^2+2^2)} = \sqrt20 = 2\sqrt5=GH$
so I think the idea is to demonstrate that it is the midpoint. I drew some auxiliary lines...
Let the vertex of right angled triangle it's hypotenuse is $GH=\sqrt 5=\sqrt{2^2+1^2}$ be M, then:
$\triangle BEF\approx \triangle MGH$
and we have:
$\frac{GN=\sqrt 5}{BF}=\frac {MN=\frac 12}{BE=1}$
which gives :
$EH=EF=2\sqrt 5$
This is true if $BF=\frac{BE}2=HI$
The ratio of sides adjacent to the rught angle is $\frac 12$ in all similar triangles MGN, BFE and HID.
From the last drawing it is easy to see that HI = GM
$\triangle HID \sim \triangle MNG \implies \frac{HI}{ID} = \frac{k}{2k}\therefore ID = 4\\ HQ=ID−GM=4−2=2.\\ \triangle HQG \sim \triangle GMN \therefore \frac{GH}{HQ}=\frac{GN}{MN}⟺\frac{GH}{2}=\frac{\sqrt5}{1} \therefore GH=EG=2\sqrt5$
Here is my interpretation of a solution to this problem.
Perform a $90^{\circ}$ clock-wise rotation around the point $D$ and let $K$ be the image of $E$ under this rotation. Then, the triangle $DCK$ is the $90^{\circ}$ clock-wise rotated image of triangle $DAE$. Therefore, $DE = DK$ and $DE \, \perp \, DK$. Let the line through point $K$ and parallel to $DE$ intersect the line through point $E$ and parallel to $DK$ at the point $L$. Then $DELK$ is a square (it is a rectangle and $DE = DK$). The points $F$ lies on $EL$ because both $EF$ and $EL$ are perpendicular to $DE$ at the same point $E$. Let $J$ be the intersection point of the lines $EF$ and $AD$. The quad $DJFK$ is a parallelogram because $JF \, || \, DK$ and $DJ \, || \, KF$. Hence $$JF = DK = EL = x + y$$ which yields $$JE = FL = y$$ Denote by $M$ the intersection point of the lines $DC$ and $KL$. Because, by the $90^{\circ}$ degree rotation around $D$, the line $DA$ is rotated to the line $DC$ and the line (not the segment) $LE$ is rotated to the line $KL$, the unique intersection point $J$ of the lines $DA$ and $LE$ is rotated the the unique intersection point $M$ of the lines $DC$ and $KL$, which means that the segment $EJ$ is rotated to the segment $KM$ and therefore $$KM = JE = FL = y$$ However, by assumption, $EF = EH = x$ because $EFGH$ is a square, which means that $$DH = DE - EH = (x + y) - x = y$$ Therefore, $$DH = y = KM \,\,\,\text{ and } \,\,\, DH \, ||\, KM$$ Consequently, $DHMK$ is a parallelogram with $\angle \, HDK = 90^{\circ}$ so it is a rectangle. Hence, $HM$ is perpendicular to $DE$. But, by assumption $HG$ is also perpendicular to $DE$ at the same point $H$, so $G$ lies on the segment $HM$. Because $EFGH$ is a square, $FLMG$ is a rectangle and hence $$MG = LF = y = KM$$ Now, if we denote by $P$ the intersection point of $BC$ and $HM$, we can deduce the fact that the triangles $KMP$ and $MGN$ are congruent because both are right-angled and $\angle \, GMN = \angle\, MKP$ (or you can argue that you can rotate $MGN$ into $KMP$ by a $90^{\circ}$ clock-wise rotation around the center of the smaller square of edge-length $y$ that they are inscribed in). Therefore, $$MP = GN = \sqrt{5}$$ Furthermore, if you look at the pair of triangles $DHI$ and $KMC$ you can see that they are also congruent (in fact they are translates of each other) because they are both right-angled, $DH = KM = y$ and $\angle \, HDI = \angle\, MKC$. Hence, $$MC = HI = 2$$ Now, by applying Pythagoras' theorem to the triangle $CMP$ we can calculate that $$CP = \sqrt{MP^2 - MC^2} = \sqrt{(\sqrt{5})^2 - 2^2} = \sqrt{5 - 4} = 1$$ But then, since the triangles $CMP$ and $MKP$ are similar $$\frac{MK}{MP} = \frac{CM}{CP} = \frac{2}{1} = 2 \,\,\,\text{ that is } \,\,\, MK = 2 \, MP = 2\, GN = 2\sqrt{5} = y$$ However, $FL = MK = 2 \sqrt{5} = 2 \, MP$ and since $MP$ and $FL$ are parallel, this is possible if and only if $M$ is the midpoint of the segment $KL$. But since $DELK$ is a square and $MH$ is perpendicular to $DE$, the point $H$ is correspondingly the midpoint of $DE$ and thus $$EH = HD = MK = 2\sqrt{5}$$
