Aleksander Twarowski from III LO Gdynia sent us his
essay on this topic.
Aurel from King Edward VI School, Southampton, has proved Part
(1) that the four triangles $r$, $s$, $t$ and $D$ have the same
area.
Can you follow some of the suggestions given for the other
parts and discover anything else?
Here is Aurel's solution:
Let the sides opposite angles $A$, $B$ and $C$ be $a$, $b$ and
$c$ respectively. The length of each side of all three squares are
therefore known.

Consider triangle $D$. Using the well known formula, the area
of triangle $D$ is given by: $$\Delta = {1\over 2}ab\sin C =
{1\over 2}bc\sin A={1\over 2}ca\sin B$$ where all three versions of
the formula are clearly equivalent because the labelling of the
triangle was arbitrary in the first place.
Consider triangle $s$. The angle $B^*$ at $B$ in this triangle
can be found by considering that the angles around a point add up
to $360^o$, so $\angle B^*=360  90 90 B = (180  B)$.
Therefore, the area of triangle $s$ is ${1\over 2}ac\sin
(180B)$.
By exactly the same reasoning the area of triangle $r$ is
${1\over 2}bc\sin (180A)$ and the area of triangle $t$ is ${1\over
2}ab\sin (180C)$.
In general we know that $\sin (180  \theta) = \sin \theta$
so:
the area of triangle $s$ is ${1\over 2}ac\sin (180B)= {1\over
2}ac\sin B$
the area of triangle $r$ is ${1\over 2}bc\sin (180A)= {1\over
2}bc\sin A$
the area of triangle $t$ is ${1\over 2}ab\sin (180C)= {1\over
2}ab\sin C$
So triangles $r$, $s$ and $t$ all have the same area as
triangle $D$ and so we have shown that all four triangles have the
same area.
The problem has also been discussed
on AskNRICH
.