Circumscribing Circle about Regular Pentagon

Theorem
About any given regular pentagon it is possible to circumscribe a circle.

Construction

 * Euclid-IV-14.png

Let $ABCDE$ be the given regular pentagon.

Bisect $\angle BCD$ and $\angle CDE$ by the straight lines $CF$ and $DF$ respectively.

Draw the circle whose center is at $F$ and whose radius is $FC$ (or $FD$).

This circle is the one required.

Proof
From $F$ join the straight lines $FB, FA, FE$.

In a similar manner to Inscribing Circle in Regular Pentagon it can be shown that $\angle CBA, \angle BAE, \angle AED$ have been bisected by the straight lines $FB, FA, FE$ respectively.

Since $\angle BCD = CDE$ and $2 \angle FCD = \angle BCD$ and $2 \angle CDF = \angle CDE$, it follows that $\angle FCD = \angle CDF$.

From Triangle with Two Equal Angles is Isosceles it follows that $FC = FD$.

Similarly it can be shown that $FB = FA = FE = FC = FD$.

Therefore the circle $ABCDE$ circumscribes the regular pentagon $ABCDE$ as required.