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.
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