To inscribe an equilateral and equiangular pentagon in a given circle.

Let *ABCDE* be the given circle.

It is required to inscribe an equilateral and equiangular pentagon in the circle *ABCDE.*

Set out the isosceles triangle *FGH* having each of the angles at *G* and *H* double the angle at *F.*

Inscribe in the circle *ABCDE* the triangle *ACD* equiangular with the triangle *FGH,* so that the angles *CAD,* *ACD,* and *CDA* equal the angles at *F,* *G,* and *H* respectively. Therefore each of the angles *ACD* and *CDA* is also double the angle *CAD.*

Now bisect the angles *ACD* and *CDA* respectively by the straight lines *CE* and *DB,* and join *AB,* *BC,* *DE,* and *EA.*

Then, since each of the angles *ACD* and *CDA* is double the angle *CAD,* and they are bisected by the straight lines *CE* and *DB,* therefore the five angles *DAC,* *ACE,* *ECD,* *CDB,* and *BDA* equal one another.

But equal angles stand on equal circumferences, therefore the five circumferences *AB,* *BC,* *CD,* *DE,* and *EA* equal one another.

But straight lines that cut off equal circumferences are equal, therefore the five straight lines *AB,* *BC,* *CD,* *DE,* and *EA* equal one another. Therefore the pentagon *ABCDE* is equilateral.

I say next that it is also equiangular.

For, since the circumference *AB* equals the circumference *DE,* add *BCD* to each, therefore the whole circumference *ABCD* equals the whole circumference *EDCB.*

And the angle *AED* stands on the circumference *ABCD,* and the angle *BAE* on the circumference *EDCB,* therefore the angle *BAE* also equals the angle *AED.*

For the same reason each of the angles *ABC,* *BCD,* and *CDE* also equals each of the angles *BAE* and *AED,* therefore the pentagon *ABCDE* is equiangular.

But it was also proved equilateral, therefore an equilateral and equiangular pentagon has been inscribed in the given circle.

Q.E.F.

The construction of this proposition is rather tedious to carry out. First, a line has to be cut according to the construction in II.11. Next, that is used in IV.10 for the construction of a 36°-72°-72° isosceles triangle. Next, that triangle is fit into the given circle using the construction IV.2. Finally, a couple more lines are drawn to finish the pentagon.
Various alternatives have have been given by others, such as Ptolemy. One of the nicest was given in 1893 by H. W. Richmond. To inscribe a regular pentagon in a circle, first draw perpendicular radii The easiest way to verify that Richmond’s construction works is by means of trigonometry. |

Draw the diagonals of the pentagram to create a regular star pentagram ACEBD inside the pentagon. These diagonals meet forming a smaller regular pentagon in the center of the original pentagon. The diagonals of that pentagon can be drawn to make an inscribed pentagram which, in turn, bound a yet smaller regular pentagon. And so forth.
For purposes of analysis, let It is evident that there are many lines parallel to the base |

There is also a series of obtuse 36°-36°-108° isosceles triangles of varying sizes.

All these parallel lines and similar triangles yield numerous relationships among the various diagonals and sides of the pentagons. Some of these relationships are additive equations:

and so forth.

Other relationships are based on the property of 36°-72°-72° triangles used in their construction in IV.10, namely that the square of the base of such a triangle equals the product of a side and the difference between the side and the base. In terms of the diagonals and sides of the pentagons, this gives the equations:

and so forth.