Proof: By Euclid
(related to Proposition: 4.11: Inscribing Regular Pentagon in Circle)
- Let the isosceles triangle $FGH$ be set up having each of the angles at $G$ and $H$ double the (angle) at $F$ [Prop. 4.10].
- And let triangle $ACD$, equiangular to $FGH$, have been inscribed in circle $ABCDE$, such that $CAD$ is equal to the angle at $F$, and the (angles) at $G$ and $H$ (are) equal to $ACD$ and $CDA$, respectively [Prop. 4.2].
- Thus, $ACD$ and $CDA$ are each double $CAD$.
- So let $ACD$ and $CDA$ have been cut in half by the straight lines $CE$ and $DB$, respectively [Prop. 1.9].
- And let $AB$, $BC$, $DE$ and $EA$ have been joined.
- Therefore, since angles $ACD$ and $CDA$ are each double $CAD$, and are cut in half by the straight lines $CE$ and $DB$, the five angles $DAC$, $ACE$, $ECD$, $CDB$, and $BDA$ are thus equal to one another.
- And equal angles stand upon equal circumferences [Prop. 3.26].
- Thus, the five circumferences $AB$, $BC$, $CD$, $DE$, and $EA$ are equal to one another [Prop. 3.29].
- Thus, the pentagon $ABCDE$ is equilateral.
- So I say that (it is) also equiangular.
- For since the circumference $AB$ is equal to the circumference $DE$, let $BCD$ have been added to both.
- Thus, the whole circumference $ABCD$ is equal to the whole circumference $EDCB$.
- And the angle $AED$ stands upon circumference $ABCD$, and angle $BAE$ upon circumference $EDCB$.
- Thus, angle $BAE$ is also equal to $AED$ [Prop. 3.27].
- So, for the same (reasons), each of the angles $ABC$, $BCD$, and $CDE$ is also equal to each of $BAE$ and $AED$.
- Thus, pentagon $ABCDE$ is equiangular.
- And it was also shown (to be) equilateral.
- Thus, an equilateral and equiangular pentagon has been inscribed in the given circle.
- (Which is) the very thing it was required to do.
∎
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References
Adapted from (subject to copyright, with kind permission)
- Fitzpatrick, Richard: Euclid's "Elements of Geometry"
