So the other day, I solved this curious mathematics puzzle using repeated applications of Pythagoras’s theorem and a little bit of algebra.

Now I realize that there is a much simpler form of the proof.

The exercise was to prove that, given two semicircles drawn into a bigger circle as shown below, the sum of the areas of the semicircles is exactly half that of the larger circle.

Again, I’m inserting a few blank lines before presenting my proof.

Once again I am labeling some vertices in the diagram for easy reference.

Our goal is to prove that the area of a circle with radius *AO* is twice the sum of the areas of two semicircles, with radii *AC* and *BD*. But that is the same as proving that the area of a circle with radius *AO* is equal to the sum of the areas of two circles, with radii *AC* and *BD*.

The *ACO*< angle is a right angle. Therefore, the area of a circle with radius *AO* is the sum of the areas of circles with radii *AC* and *CO*. (To see this, just multiply the theorem of Pythagoras by π.) So if only we could prove that *CO* = *BD*, our proof would be complete.

Since *AO* = *BO*, they are the sides of the isosceles triangle *ABO*. Now if we were to pick a point *O*‘ on the line *CD* such that *CO*‘ = *BD*, the *ACO*‘ and *O*‘*DB* triangles will be identical (*CD* being the sum of *AC* and *BD* by construction). Therefore, *AO*‘ = *BO*‘, and the *ABO*‘ triangle would be another isosceles triangle with its third vertex on the *CD* line. Clearly that is not possible, so *O* = *O*‘, and therefore, *CO* = *BD*. This concludes the proof.