Csdrhrt

We will, first of all, prove a very interesting property

$mathbf{Lemma;1}$

Given two squares PQRS and PTUV (as shown on the picture), the triangles $Delta STP$ and $Delta PVQ$ have equal area.

$mathbf {Proof}$

Denote by $alpha$ the angle SPT and by $[...]$ the area of the polygon "...". Hence
$$[Delta STP]=frac{overline {PS}cdotoverline {PT}cdot sin(alpha)}{2}$$ $$[Delta PVQ]=frac{overline {QP}*overline {PV}cdotsinBigl(360°-(90°+90+alpha)Bigr)}{2}=frac{overline {QP}cdotoverline {PV}cdotsinBigl(180°-alphaBigr)}{2}=frac{overline {QP}cdotoverline {PV}cdotsin(alpha)}{2}$$

Since $overline {PS}=overline {PQ}$ and $overline {PT}=overline {PV}$ $$[Delta STP]=[Delta PVQ]$$

Now, back to the problem

Let $overline {AB}=a$ and $overline {IJ}=b$. Note first of all that $$Delta BEC cong Delta EIF$$
See why? $mathbf {Hint:}$

It is obvious that $overline {CE}=overline {EF}$. Use the properties of right triangles in order to show that all angles are equal.

Thus $${(overline{CE})^2}={a^2}+{b^2}=S$$

Note furthermore that $$[Delta BEC]=[Delta EIF]=frac{ab}{2}$$
By Lemma 1:
$$[Delta DCG]=[Delta BEC]=frac{ab}{2}=[Delta EIF]=[Delta GFK]$$
The area of the polygon AJKGD is thus
$$[AJKGD]=[ABCD]+[CEFG]+[FIJK]+4[Delta DCG]=2Bigl({a^2}+{b^2}Bigr)+2ab$$

The area of the trapezoid AJKD is moreover
$$[AJKD]=frac{(a+b)(2a+2b)}{2}={a^2}+2ab+{b^2}$$

Finally
$$T=[Delta DKG]=[AJKGD]-[AJKD]={a^2}+{b^2}=S Rightarrow S=T$$

$$|square P_1 P_2 P_3 P_4| = (a+b)^2 = frac12(a+b)(2a+2b) = |square Q_1 Q_2 Q_3 Q_4|quad=:R$$

$$S ;=; R - 4cdotfrac12ab ;=; T$$

(This space intentionally left blank.)

Because there are so many squares, coordinates are easy to compute.

The area of the shaded square is clearly $u^2+v^2$.

The area of the shaded triangle is one-half of the absolute value of the determinant of the array

$$left[ begin{array}{c}
1 & 1 & 1 \
2u-v & 3u & 2u \
3u+v & u+3v & u+v
end{array} right]$$

which is also $u^2+v^2$.

I have a second solution.

$triangle GPN$ is obtained by rotating $triangle GSD 90^circ$ clockwise.
$triangle GQM$ is obtained by rotating $triangle GRK 90^circ$ counterclockwise.

The four triangles adjacent to $S$ (two of them right, two of them obtuse) all have the same area. (Each has the same base and height as the one on the opposite side of the square, while the two right triangles are congruent.). Now rotate each of the obtuse triangles by $90^circ$ so that they are adjacent to $T$, as shown.

What now needs to be proved is that the two shaded pentagons have equal area. This can be done by observing that each pentagon decomposes into an isosceles right triangle and a trapezoid. The isosceles right triangles are congruent; the trapezoids have equal area as their two bases are the same and their heights are the same.

While the other solutions are obviously correct, they are also unnecessarily complicated.

Since the angle of the squares is not specified, it must be true for all angles, so* why not pick one which is simple to work with and results in a degenerate case.

*) The assumption of truth is not required since we first do show that S=T is, in fact, true (in one simple case) and skirt the rules from there on by leaving the extrapolation to the reader.

While this approach is best suited for puzzles, looking at the edge cases first is a quick way to disprove things by example, or at least check that your calculations are correct