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Stage: 5 Challenge Level: Challenge Level:3 Challenge Level:3 Challenge Level:3

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 ${1\over 2}ac\sin (180-B)$.

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

In general we know that $\sin (180 - \theta) = \sin \theta$ so:

the area of triangle $s$ is ${1\over 2}ac\sin (180-B)= {1\over 2}ac\sin B$
the area of triangle $r$ is ${1\over 2}bc\sin (180-A)= {1\over 2}bc\sin A$
the area of triangle $t$ is ${1\over 2}ab\sin (180-C)= {1\over 2}ab\sin C$

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.