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 triangleTand the area of squareSare 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.

**Answer**

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

Lemma1

Given two squares PQRS and PTUV (as shown on the picture), the triangles ΔSTP and ΔPVQ have equal area.

Proof

Denote by α the angle SPT and by [...] the area of the polygon “…”. Hence

[ΔSTP]=¯PS⋅¯PT⋅sin(α)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

**Attribution***Source : Link , Question Author : Larry , Answer Author : Community*