Here are somewhat-natural generalizations of a couple of the two-squares-joined-at-a-vertex results from OP's answer.

• Potema's Theorem

Let squares $\square A'B'C'D'$ and $\square A''B''C''D''$ be as shown, with $X$ the midpoint of $X'X''$.

If segment $A'A''$ remains fixed, and segment $D'D''$ keeps a constant length and inclination (in Potema's theorem, $D'=D''$), then midpoint $B$ remains fixed. Moreover, the position of $C$ relative to $D$ matches that of $B$ relative to $A$.

That is, $\square ABCD$ is a parallelogram, with $|AB|$ and $|CD|$ determined by the lengths and relative inclinations of $A'A''$ and $C'C''$.

Specifically, segments $AB$ and $CD$ form congruent triangles with segments of length $|AA'|$ and $|DD'|$ (more-specifically, segments perpendicular to $A'A''$ and $B'B''$). Likewise, segments $BC$ and $AD$ form congruent triangles with segments of length $|BB'|$ and $|CC'|$.

• Fensler-Hadwiger

Again, we have squares $\square A'B'C'D'$ and $\square A''B''C''D''$, and midpoints $X$ of $X'X''$.

The square with opposite vertices at the centers of the two squares has its other two vertices equally offset from $B$ and $D$: namely, by distance $|AA'|$ and in directions perpendicular to $A'A''$. (Moreover, they're offset by $|CC'|$ in directions perpendicular to $C'C''$.)

In Fensler-Hedwiger, $A'=A''$, so that the new square's vertices are $B$ and $D$ themselves. Not mentioned in OP's statement of FH is that the diameter of the square is equal to $|CC'|$, and that $BD\perp C'C''$.

There's a bonus square with opposite vertices $A$ and $C$; the remaining vertices are midpoints of $B'D''$ and $B''D'$. This is actually an instance of a deeper theorem: If $P_0P_1\cdots P_n$ and $Q_0Q_1\cdots Q_n$ are similar polygonal paths with opposite orientations in plane (that is, if they're related by a dilation with a negative scale factor), then the midpoints of $P_iQ_i$ form another similar path.

Other sets of midpoints form parallelograms.

• $A_1+A_2 = \frac12(M^2+N^2)$

Let squares $\square A'B'C'D'$ and $\square A''B''C''D''$ be as shown, and define $x := |X'X''|$. Writing $S'$ and $S''$ for their areas, and $S$ for the area of the square with opposite vertices at their centers, we can find that

$$a^2+b^2+c^2+d^2 \;=\; 2 \left(\; S' + S'' + 4 S \;\right)$$

For the referenced result from OP's list, we can take $A'=A''$, so that $a=0$. From comments above about the Fensler-Hadwiger theorem, we also have that the diameter of the center square is $\frac12c$. Hence, $S = \frac18 c^2$, and the generalized result specializes back to $b^2+d^2=2(S'+S'')$.

Here are somewhat-natural generalizations of a couple of the two-squares-joined-at-a-vertex results from OP's answer.

• Potema's Theorem

Let squares $\square A'B'C'D'$ and $\square A''B''C''D''$ be as shown, with $X$ the midpoint of $X'X''$.

If segment $A'A''$ remains fixed, and segment $D'D''$ keeps a constant length and inclination (in Potema's theorem, $D'=D''$), then midpoint $B$ remains fixed. Moreover, the position of $C$ relative to $D$ matches that of $B$ relative to $A$.

That is, $\square ABCD$ is a parallelogram, with $|AB|$ and $|CD|$ determined by the lengths and relative inclinations of $A'A''$ and $C'C''$.

Specifically, segments $AB$ and $CD$ form congruent triangles with segments of length $|AA'|$ and $|DD'|$ (more-specifically, segments perpendicular to $A'A''$ and $B'B''$). Likewise, segments $BC$ and $AD$ form congruent triangles with segments of length $|BB'|$ and $|CC'|$.

• Fensler-Hadwiger

Again, we have squares $\square A'B'C'D'$ and $\square A''B''C''D''$, and midpoints $X$ of $X'X''$.

The square with opposite vertices at the centers of the two squares has its other two vertices equally offset from $B$ and $D$: namely, by distance $|AA'|$ and in directions perpendicular to $A'A''$. (Moreover, they're offset by $|CC'|$ in directions perpendicular to $C'C''$.)

In Fensler-Hedwiger, $A'=A''$, so that the new square's vertices are $B$ and $D$ themselves. Not mentioned in OP's statement of FH is that $BD\perp C'C''$; also, the diameter of the square is equal to $|CC'|$, but this fact is obvious:

There's a bonus square with opposite vertices $A$ and $C$; the remaining vertices are midpoints of $B'D''$ and $B''D'$. This is actually an instance of a deeper theorem: If $P_0P_1\cdots P_n$ and $Q_0Q_1\cdots Q_n$ are similar polygonal paths with opposite orientations in plane (that is, if they're related by a dilation with a negative scale factor), then the midpoints of $P_iQ_i$ form another similar path.

Other sets of midpoints form parallelograms.

• $A_1+A_2 = \frac12(M^2+N^2)$

Let squares $\square A'B'C'D'$ and $\square A''B''C''D''$ be as shown, and define $x := |X'X''|$. Writing $S'$ and $S''$ for their areas, and $S$ for the area of the square with opposite vertices at their centers, we can find that

$$a^2+b^2+c^2+d^2 \;=\; 2 \left(\; S' + S'' + 4 S \;\right)$$

For the referenced result from OP's list, we can take $A'=A''$, so that $a=0$. From comments above about the Fensler-Hadwiger theorem, we also have that the diameter of the center square is $\frac12c$. Hence, $S = \frac18 c^2$, and the generalized result specializes back to $b^2+d^2=2(S'+S'')$.

Here are somewhat-natural generalizations of a couple of the two-squares-joined-at-a-vertex results from OP's answer.

• Potema's Theorem

Let squares $\square A'B'C'D'$ and $\square A''B''C''D''$ be as shown, with $X$ the midpoint of $X'X''$.

If segment $A'A''$ remains fixed, and segment $D'D''$ keeps a constant length and inclination (in Potema's theorem, $D'=D''$), then midpoint $B$ remains fixed. Moreover, the position of $C$ relative to $D$ matches that of $B$ relative to $A$.

That is, $\square ABCD$ is a parallelogram, with $|AB|$ and $|CD|$ determined by the lengths and relative inclinations of $A'A''$ and $C'C''$.

Specifically, segments $AB$ and $CD$ form congruent triangles with segments of length $|AA'|$ and $|DD'|$ (more-specifically, segments perpendicular to $A'A''$ and $B'B''$). Likewise, segments $BC$ and $AD$ form congruent triangles with segments of length $|BB'|$ and $|CC'|$.

• Fensler-Hadwiger

Again, we have squares $\square A'B'C'D'$ and $\square A''B''C''D''$, and midpoints $X$ of $X'X''$.

The square with opposite vertices at the centers of the two squares has its other two vertices equally offset from $B$ and $D$: namely, by distance $|AA'|$ and in directions perpendicular to $A'A''$. (Moreover, they're offset by $|CC'|$ in directions perpendicular to $C'C''$.)

In Fensler-Hedwiger, $A'=A''$, so that the new square's vertices are $B$ and $D$ themselves. Not mentioned in OP's statement of FH is that the diameter of the square is equal to $|CC'|$, and that $BD\perp C'C''$.

There's a bonus square with opposite vertices $A$ and $C$; the remaining vertices are midpoints of $B'D''$ and $B''D'$. This is actually an instance of a deeper theorem: If $P_0P_1\cdots P_n$ and $Q_0Q_1\cdots Q_n$ are similar polygonal paths with opposite orientations in plane (that is, if they're related by a dilation with a negative scale factor), then the midpoints of $P_iQ_i$ form another similar path.

Other sets of midpoints form parallelograms.

Here are somewhat-natural generalizations of a couple of the two-squares-joined-at-a-vertex results from OP's answer.

• Potema's Theorem

Let squares $\square A'B'C'D'$ and $\square A''B''C''D''$ be as shown, with $X$ the midpoint of $X'X''$.

If segment $A'A''$ remains fixed, and segment $D'D''$ keeps a constant length and inclination (in Potema's theorem, $D'=D''$), then midpoint $B$ remains fixed. Moreover, the position of $C$ relative to $D$ matches that of $B$ relative to $A$.

That is, $\square ABCD$ is a parallelogram, with $|AB|$ and $|CD|$ determined by the lengths and relative inclinations of $A'A''$ and $C'C''$.

Specifically, segments $AB$ and $CD$ form congruent triangles with segments of length $|AA'|$ and $|DD'|$ (more-specifically, segments perpendicular to $A'A''$ and $B'B''$). Likewise, segments $BC$ and $AD$ form congruent triangles with segments of length $|BB'|$ and $|CC'|$.

• Fensler-Hadwiger

Again, we have squares $\square A'B'C'D'$ and $\square A''B''C''D''$, and midpoints $X$ of $X'X''$.

The square with opposite vertices at the centers of the two squares has its other two vertices equally offset from $B$ and $D$: namely, by distance $|AA'|$ and in directions perpendicular to $A'A''$. (Moreover, they're offset by $|CC'|$ in directions perpendicular to $C'C''$.)

In Fensler-Hedwiger, $A'=A''$, so that the new square's vertices are $B$ and $D$ themselves. Not mentioned in OP's statement of FH is that the diameter of the square is equal to $|CC'|$, and that $BD\perp C'C''$.

There's a bonus square with opposite vertices $A$ and $C$; the remaining vertices are midpoints of $B'D''$ and $B''D'$. This is actually an instance of a deeper theorem: If $P_0P_1\cdots P_n$ and $Q_0Q_1\cdots Q_n$ are similar polygonal paths with opposite orientations in plane (that is, if they're related by a dilation with a negative scale factor), then the midpoints of $P_iQ_i$ form another similar path.

Other sets of midpoints form parallelograms.

• $A_1+A_2 = \frac12(M^2+N^2)$

Let squares $\square A'B'C'D'$ and $\square A''B''C''D''$ be as shown, and define $x := |X'X''|$. Writing $S'$ and $S''$ for their areas, and $S$ for the area of the square with opposite vertices at their centers, we can find that

$$a^2+b^2+c^2+d^2 \;=\; 2 \left(\; S' + S'' + 4 S \;\right)$$

For the referenced result from OP's list, we can take $A'=A''$, so that $a=0$. From comments above about the Fensler-Hadwiger theorem, we also have that the diameter of the center square is $\frac12c$. Hence, $S = \frac18 c^2$, and the generalized result specializes back to $b^2+d^2=2(S'+S'')$.