Prove that a regular pentagon is cyclic

Question

Prove that a regular pentagon is cyclic

I tried to use that every isosceles trapezoid is cyclic, so if you make a segment of some vertices that are not collinear you get an isosceles trapezoid and a triangle and both are cyclic, but I don't know how to relate them.

Answer 1

Let $ABCDE$ be our pentagon.

Thus, by your work ($ABCD$ and $ABCE$ are cyclics) $D$ and $E$ are placed on the circle $ABC$.

Answer 2

Let $ABCDE$ be your regular pentagon. Let $\gamma$ be the circle passing through $ABC$, and let $O$ be the center of $\gamma$.

First, you can prove that the triangles $ABO$ and $BCO$ are congruent, because they have the same sides.

Then you can prove that the triangles $BCO$ and $CDO$ are congruent, using the fact that $$BC \cong CD$$ $$BO \cong CO$$ $$\widehat{OBC}=\widehat{ABC}-\widehat{ABO} \cong \widehat{BCD}-\widehat{BCO} =\widehat{OCD}$$

Thus $OD \cong OC$, and you proved that $D \in \gamma$.

In a similar way you can prove that $E \in \gamma$ as well, the proof repeats using the triangles $CDO$ and $DEO$.

EDIT: The same proof applies to any $N$-sided regular polygon $A_1 A_2 \ldots A_N$. You fix the circle $\gamma$ passing through $A_1, A_2, A_3$ and prove that $A_4 \in \gamma$. This means that $\gamma$ is the circle through $A_2, A_3, A_4$, and you prove that $A_5 \in \gamma$. Then repeat recursivley until you get $A_N \in \gamma$.

Answer 3

All regular polygons are cyclic. That's almost the definition.

Equiangular vertices and equal sides if and only if circumscribed and equal distance apart.

Because:

Take a side between two adjacent vertices $A,B$. The angles $\angle A$ and $\angle B$ are equal because the polygon is regular. The measure of the angles are less then $180$ and in pentagon they are $108$. Construct angle bisectors of $\angle A$ and $\angle B$. And the bisecteed angles are acute (in this case $54$ degrees) the bisectors will intersect at a point $O$ and $\triangle AOB$ is isosceles.

Now take the side between the adjacent vertices $B,C$ and bisect $\angle C$. The $P$ be the point where the angle bisector of $C$ intersects the angle bisector of $B$. Then $\triangle BPC$ is isosceles. And $AB = BC$ and as $\angle B \cong \angle C$ the base angles of the triangles are equal and the triangles are congruent. So $BP = BO$ and $O=P$ and $O$ is the intersection of the the angle bisectors.

Do this of all sides and we conclude all angle bisectors intersect at $O$ and the distance from any vertex to $O$ is equal. Thus circumscribed.