To calculate the areas take $AB = BC = CD = 2$ units.

The circle with $AD$ as diameter has area $9\pi$ (radius $3$).

A circle with $AB$ as diameter would have area $\pi$.

A circle with $AC$ as diameter would have area $4\pi$.

The unshaded area is $(2 \times2\pi) - (2 \times\pi/2) = 3\pi$.

The total shaded area is $6\pi$ and from the symmetry of the diagram it can be seen that this is divided into two equal portions of area $3\pi$.

David of Madras College, St Andrew's had a novel way of looking at this. He first rearranged the semicircles into three circles touching at $A$. He then pointed out that, when working out the areas, all the answers will be multiples of $\pi$ and ultimately, when comparing the areas, it is necessary to factorise out $\pi$ anyway. David's geometrical representation of the results "when divided by $\pi$'', compares the equivalent areas of squares.

Thomas of Simon Langton School, Canterbury also sent in a good solution.