Japanese Temple Problem From 1844 I recently learnt a Japanese geometry temple problem. 
The problem is the following:
Five squares are arranged as the image shows. Prove that the area of triangle T and the area of  square S are equal.

This is problem 6 in this article.
I am thinking about law of cosines, but I have not been able to prove the theorem. Any hints would be appreciated. 
 A: 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
$$[\Delta STP]=\frac{\overline {PS}\cdot\overline {PT}\cdot \sin(\alpha)}{2}$$ $$[\Delta PVQ]=\frac{\overline {QP}*\overline {PV}\cdot\sin\Bigl(360°-(90°+90+\alpha)\Bigr)}{2}=\frac{\overline {QP}\cdot\overline {PV}\cdot\sin\Bigl(180°-\alpha\Bigr)}{2}=\frac{\overline {QP}\cdot\overline {PV}\cdot\sin(\alpha)}{2}$$
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
$$[AJKGD]=[ABCD]+[CEFG]+[FIJK]+4[\Delta DCG]=2\Bigl({a^2}+{b^2}\Bigr)+2ab$$
The area of the trapezoid AJKD is moreover
$$[AJKD]=\frac{(a+b)(2a+2b)}{2}={a^2}+2ab+{b^2}$$
Finally
$$T=[\Delta DKG]=[AJKGD]-[AJKD]={a^2}+{b^2}=S \Rightarrow S=T$$
A: 
$$|\square P_1 P_2 P_3 P_4| = (a+b)^2 = \frac12(a+b)(2a+2b) = |\square Q_1 Q_2 Q_3 Q_4|\quad=:R$$

$$S \;=\; R - 4\cdot\frac12ab \;=\; T$$

(This space intentionally left blank.)
A: 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.
A: 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
A: 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.
A: This is a long comment.
The shapes $S,\,T$ share a vertex $A$ where six angles meet, three of them right angles. The other three angles therefore sum to a right angle. Let $\theta$ be the angle in $T$, so the two other acute angles, and the right angle between them from $S$, sum to $\pi-\theta$. Since $\sin\theta=\sin(\pi-\theta)$, $T$ has the same area as $\triangle ABC$, where $BA,\,AC$ are sides of the upper squares.
The challenge, then, is to show $S$ has that much area too. Let $D$ denote the vertex of $S$ opposite $A$. I suspect we can show the quadrilateral $BACD$ has twice the area of $S$, and also twice the area of $\triangle ABC$.
