Notes
three squares in a semi-circle solution

Solution to the Three Squares in a Semi-Circle Puzzle

Solution by Pythagoras' Theorem

Let $a$ be the side length of the smallest squares, so that $a^2 = 6$, and let $b$ be the side length of the larger one. As the perpendicular bisector of a chord passes through the centre of the circle, the continuation of $D E$ passes through the centre of the circle at $O$.

Triangle $B D C$ is right-angled with $B D$ of length $a$ and $D C$ of length $b - 2 a$. The desired length, $B C$, is the hypotenuse of this triangle so by Pythagoras' theorem the square of its length is given by:

Let $r$ be the radius of the circle. Triangle $B D O$ is right-angled with side lengths $a$, $b + a$, and $r$, so by Pythagoras' theorem:

Triangle $O E C$ is also right-angled and its side lengths are $b$, $b - a$, and $r$ so:

Subtracting the second equation from the first gives:

Adding this to the equation for the square of the length of $C B$ gives that that quantity is $6 a^2$ which is $36$. Therefore, the length of $C B$ is $6$.