| 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. | IV.10 | |
| Now bisect the angles ACD and CDA respectively by the straight lines CE and DB, and join AB, BC, DE, and EA. | I.9 | |
| 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. | III.26 | |
| 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. | III.29 | |
| 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. | III.27 | |
| 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. |
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| Q.E.F. | ||
| Euclid's construction implicit in this proposition is rather tedious to carry out. Various alternatives have have been given by others, such as Ptolemy. In 1893 H. W. Richmond gave the following construction for inscribing a regular pentagon in a circle. Draw perpendicular radii OA and OB from the center O of a circle. Let C be the midpoint of OB and draw AC. Bisect angle ACO to meet OA at D. Draw a perpendicular DE to OA to the circle. Then AE is one side of the pentagon. The remaining sides can then be constructed. |
Book IV Introduction - Proposition IV.10 - Proposition IV.12.