Construction of Equilateral Polygon with Even Number of Sides in Outer of Concentric Circles

Proof

 * Euclid-XII-16.png

Let $ABCD$ and $EFGH$ be two concentric circles whose centers are at $K$.

Let $ABCD$ be the larger.

It is required that a regular polygon with an even number of sides be inscribed in $ABCD$ such that it does not touch $EFGH$.

Let the straight line $BKD$ be drawn through the center $K$.

From the point $G$ let $GA$ be drawn perpendicular to $BD$ and carried through to $ABCD$ at $C$.

From :
 * $AC$ is tangent to $EFGH$.

Let the arc $BAD$ of $ABCD$ be bisected repeatedly.

By :
 * this can be done until an arc remains less than $AD$.

Let such remaining arc be $LD$.

Let $LM$ be drawn perpendicular to $BD$ and carried through to $ABCD$ at $N$.

Let $LD$ and $DN$ be joined.

From
 * $LD = DN$

We have that $LN \parallel AC$.

We also have that $AC$ is tangent to $EFGH$.

Therefore $LN$ does not touch $EFGH$.

Therefore $LD$ and $DN$ are far from touching $EFGH$.

We can then fit straight lines equal to $LD$ into $ABCD$ as chords going all the way round the circle.

Thus a regular polygon with an even number of sides has been inscribed in $ABCD$ such that it does not touch $EFGH$.