If in an equilateral and equiangular pentagon straight lines
subtend two angles are taken in order, then they cut one another in extreme and mean ratio, and their greater segments equal the side of the pentagon.

In the equilateral and equiangular pentagon *ABCDE* let the straight lines *AC* and *BE,* cutting one another at the point *H,* subtend two angles taken in order, the angles at *A* and *B.*

I say that each of them has been cut in extreme and mean ratio at the point *H,* and their greater segments equal the side of the pentagon.

Circumscribe the circle *ABCDE* about the pentagon *ABCDE.*

Then, since the two straight lines *EA* and *AB* equal the two lines *AB* and *BC,* and they contain equal angles, therefore the base *BE* equals the base *AC,* the triangle *ABE* equals the triangle *ABC,* and the remaining angles equal the remaining angles respectively, namely those opposite the equal sides.

Therefore the angle *BAC* equals the angle *ABE.* Therefore the angle *AHE* is double the angle *BAH.*

But the angle *EAC* is also double the angle *BAC,* for the circumference *EDC* is also double the circumference *CB.*

Therefore the angle *HAE* equals the angle *AHE.* Hence the straight line *HE* also equals *EA,* that is, *AB.*

And, since the straight line *BA* equals *AE,* therefore the angle *ABE* also equals the angle *AEB.*

But the angle *ABE* was proved equal to the angle *BAH,* therefore the angle *BEA* also equals the angle *BAH.*

And the angle *ABE* is common to the two triangles *ABE* and *ABH,* therefore the remaining angle *BAE* equals the remaining angle *AHB.* Therefore the triangle *ABE* is equiangular with the triangle *ABH.*

Therefore, proportionally *EB* is to *BA* as *AB* is to *BH.*

But *BA* equals *EH,* therefore *BE* is to *EH* as *EH* is to *HB.*

And *BE* is greater than *EH,* therefore *EH* is also greater than *HB.*

Therefore *BE* has been cut in extreme and mean ratio at *H,* and the greater segment *HE* equals the side of the pentagon.

Similarly we can prove that *AC* has also been cut in extreme and mean ratio at *H,* and its greater segment *CH* equals the side of the pentagon.

Therefore, *if in an equilateral and equiangular pentagon straight lines subtend two angles are taken in order, then they cut one another in extreme and mean ratio, and their
greater segments equal the side of the pentagon.*

Q.E.D.

This proposition shows why cutting a line in extreme and mean ratio is so important.

This proposition is used in the proof of XIII.11 to establish that the side of a regular pentagon inscribed in a circle with rational diameter is the irrational straight line called minor.