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:

D =

absinC =

bcsinA=

casinB

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 ÐB^*=360 - 90 -90- B = (180 - B).

Therefore, the area of triangle s is ¹/₂acsin(180-B).

By exactly the same reasoning the area of triangle r is ¹/₂bcsin(180-A) and the area of triangle t is ¹/₂absin(180-C).

In general we know that sin(180 - q) = sinq so:
the area of triangle s is ¹/₂acsin(180-B) = ¹/₂acsinB
the area of triangle r is ¹/₂bcsin(180-A) = ¹/₂bcsinA
the area of triangle t is ¹/₂absin(180-C) = ¹/₂absinC

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 .