- Let $ABCDE$ be a circle.
- And let the equilateral pentagon $ABCDE$ have been inscribed in circle $ABCDE$.
- I say that the square on the side of pentagon $ABCDE$ is the (sum of the squares) on the sides of the hexagon and of the decagon inscribed in circle $ABCDE$.

- For let the center of the circle, point $F$, have been found [Prop. 3.1].
- And, $AF$ being joined, let it have been drawn across to point $G$.
- And let $FB$ have been joined.
- And let $FH$ have been drawn from $F$ perpendicular to $AB$.
- And let it have been drawn across to $K$.
- And let $AK$ and $KB$ have been joined.
- And, again, let $FL$ have been drawn from $F$ perpendicular to $AK$.
- And let it have been drawn across to $M$.
- And let $KN$ have been joined.
- Since circumference $ABCG$ is equal to circumference $AEDG$, of which $ABC$ is equal to $AED$, the remaining circumference $CG$ is thus equal to the remaining (circumference) $GD$.
- And $CD$ (is the side) of the pentagon.
- $CG$ (is) thus (the side) of the decagon.
- And since $FA$ is equal to $FB$, and $FH$ is perpendicular (to $AB$), angle $AFK$ (is) thus also equal to $KFB$ [Prop. 1.5], [Prop. 1.26].
- Hence, circumference $AK$ is also equal to $KB$ [Prop. 3.26].
- Thus, circumference $AB$ (is) double circumference $BK$.
- Thus, straight line $AK$ is the side of the decagon.
- So, for the same (reasons, circumference) $AK$ is also double $KM$.
- And since circumference $AB$ is double circumference $BK$, and circumference $CD$ (is) equal to circumference $AB$, circumference $CD$ (is) thus also double circumference $BK$.
- And circumference $CD$ is also double $CG$.
- Thus, circumference $CG$ (is) equal to circumference $BK$.
- But, $BK$ is double $KM$, since $KA$ (is) also (double $KM$).
- Thus, (circumference) $CG$ is also double $KM$.
- But, indeed, circumference $CB$ is also double circumference $BK$.
- For circumference $CB$ (is) equal to $BA$.
- Thus, the whole circumference $GB$ is also double $BM$.
- Hence, angle $GFB$ [is] also double angle $BFM$ [Prop. 6.33].
- And $GFB$ (is) also double $FAB$.
- For $FAB$ (is) equal to $ABF$.
- Thus, $BFN$ is also equal to $FAB$.
- And angle $ABF$ (is) common to the two triangles $ABF$ and $BFN$.
- Thus, the remaining (angle) $AFB$ is equal to the remaining (angle) $BNF$ [Prop. 1.32].
- Thus, triangle $ABF$ is equiangular to triangle $BFN$.
- Thus, proportionally, as straight line $AB$ (is) to $BF$, so $FB$ (is) to $BN$ [Prop. 6.4].
- Thus, the (rectangle contained) by $ABN$ is equal to the (square) on $BF$ [Prop. 6.17].
- Again, since $AL$ is equal to $LK$, and $LN$ is common and at right angles (to $KA$), base $KN$ is thus equal to base $AN$ [Prop. 1.4].
- And, Thus, angle $LKN$ is equal to angle $LAN$.
- But, $LAN$ is equal to $KBN$ [Prop. 3.29], [Prop. 1.5].
- Thus, $LKN$ is also equal to $KBN$.
- And the (angle) at $A$ (is) common to the two triangles $AKB$ and $AKN$.
- Thus, the remaining (angle) $AKB$ is equal to the remaining (angle) $KNA$ [Prop. 1.32].
- Thus, triangle $KBA$ is equiangular to triangle $KNA$.
- Thus, proportionally, as straight line $BA$ is to $AK$, so $KA$ (is) to $AN$ [Prop. 6.4].
- Thus, the (rectangle contained) by $BAN$ is equal to the (square) on $AK$ [Prop. 6.17].
- And the (rectangle contained) by $ABN$ was also shown (to be) equal to the (square) on $BF$.
- Thus, the (rectangle contained) by $ABN$ plus the (rectangle contained) by $BAN$, which is the (square) on $BA$ [Prop. 2.2], is equal to the (square) on $BF$ plus the (square) on $AK$.
- And $BA$ is the side of the pentagon, and $BF$ (the side) of the hexagon [Prop. 4.15 corr.] 0, and $AK$ (the side) of the decagon.
- Thus, the square on the side of the pentagon (inscribed in a circle) is (equal to) the (sum of the squares) on the (sides) of the hexagon and of the decagon inscribed in the same circle.∎

**Fitzpatrick, Richard**: Euclid's "Elements of Geometry"