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:

Δ = \frac{1}{2} ab \sin C = \frac{1}{2} bc \sin A = \frac{1}{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 $∠ B^{*} = 360 - 90 - 90 - B = (180 - B)$ .

Therefore, the area of triangle $s$ is $\frac{1}{2} ac \sin (180 - B)$ .

By exactly the same reasoning the area of triangle $r$ is $\frac{1}{2} bc \sin (180 - A)$ and the area of triangle $t$ is $\frac{1}{2} ab \sin (180 - C)$ .

In general we know that $\sin (180 - θ) = \sin θ$ so:
the area of triangle $s$ is $\frac{1}{2} ac \sin (180 - B) = \frac{1}{2} ac \sin B$
the area of triangle $r$ is $\frac{1}{2} bc \sin (180 - A) = \frac{1}{2} bc \sin A$
the area of triangle $t$ is $\frac{1}{2} ab \sin (180 - C) = \frac{1}{2} ab \sin C$

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 .