In the Math Corner of my May Diary I included the following brainteaser.
I have parked two semicircles inside a circle, as shown. What looks to be the case, is the case: AB and CD are parallel, E is a single point of contact.
Prove that, within the larger circle, the shaded area is equal to the non-shaded area; in other words, that the sum of the areas of the two semicircles is half the area of the big circle.
There are three circles in play here. There's the smallest one, of which AEB is a semicircle; then a slightly bigger one, of which CED is a semicircle; and then the big enclosing circle. I'll call the radii of these three circles p, q, and r. Just remember the order, smallest to largest: p, q, and r.
Applying the formula for the area of a circle, we have to prove that πp²/2 + πq²/2 = πr²/2, which of course simplifies to p² + q² = r². Aha! — Pythagoras' theorem. But where is the right-angled triangle?
First construct a diameter of the big circle parallel to AD. I've drawn it in red here. Now imagine that diameter is a mirror and reflect the line CD in the mirror.
C reflects to C′, which plainly is just the same as B. D reflects to some point on the big circle's circumference a wee bit clockwise of D — call it D′; and it is easy to see that D′B is perpendicular to AB, 45° plus 45°.
There's your right-angled triangle, ABD′. And guess what: AD′ = 2r (because any chord that subtends a right angle at the circle's circumference must be a diameter), BD′ = 2q (because it's the reflection of CD), and AB = 2p (duh). The result follows from Pythagoras' Theorem.
Added later: A reader sent in a niftier solution. From the center of the big circle, draw raddi to A and C. We then have:
Now add first and last equations and square!