Japanese Temple Problem From 1844

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