Fitting Chord Into Circle

Theorem
Into a given circle, it is possible to fit a chord equal to a given straight line which is not greater than the diameter of the circle.

Construction

 * Euclid-IV-1.png

Let $ABC$ be the given circle and $D$ be the given straight line.

Let $BC$ be a diameter of circle $ABC$.

If $D = BC$ then the job is done, as $BC$ is already fitted into $ABC$.

Otherwise, let $CE$ be cut off from $BC$ equal to $D$.

With center $C$ and radius $CE$, draw circle $EFG$.

Let $G$ be one of the points at which $EFG$ meets $ABC$.

Then $CG$ is the chord required.

Proof
As $C$ is the center of $EFG$, $CG = CE$ and so $CG = D$.

But $CG$ is a chord of $ABC$.