Btukfyl

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.

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
$$[Delta STP]=frac{overline {PS}*overline {PT}*sin(alpha)}{2}$$ $$[Delta PVQ]=frac{overline {QP}*overline {PV}*sinBigl(360°-(90°+90+alpha)Bigr)}{2}=frac{overline {QP}*overline {PV}*sinBigl(180°-alphaBigr)}{2}=frac{overline {QP}*overline {PV}*sin(alpha)}{2}$$

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
$$[AJKGD]=[ABCD]+[CEFG]+[FIJK]+4[Delta DCG]=2Bigl({a^2}+{b^2}Bigr)+2ab$$

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

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

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 $sintheta=sin(pi-theta)$, $T$ 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 $triangle ABC$.