If the side of the hexagon and that of the decagon inscribed in the same circle are added together, then the whole straight line has been cut in extreme and mean ratio, and its greater segment is the side of the hexagon.

Let *ABC* be a circle, and of the figures inscribed in the circle *ABC* let *BC* be the side of a decagon, and *CD* that of a hexagon, and let them be in a straight line.

I say that the whole straight line *BD* is cut in extreme and mean ratio, and *CD* is its greater segment.

Take the center *E* of the circle, join *EB, EC,* and *ED,* and carry *BE* through to *A.*

Since *BC* is the side of an equilateral decagon, therefore the circumference *ACB* is five times the circumference *BC.* Therefore the circumference *AC* is quadruple *CB.*

But the circumference *AC* is to *CB* as the angle *AEC* is to the angle *CEB.* Therefore the angle *AEC* is quadruple the angle *CEB.*

And, since the angle *EBC* equals the angle *ECB,* therefore the angle *AEC* is double the angle *ECB.*

And, since the straight line *EC* equals *CD,* for each of them equals the side of the hexagon inscribed in the circle *ABC.* Therefore the angle *CED* also equals the angle *CDE.* Therefore the angle *ECB* is double the angle *EDC.*

But the angle *AEC* was proved double the angle *ECB,* therefore the angle *AEC* is quadruple the angle *EDC.* And the angle *AEC* was also proved quadruple the angle *BEC,* therefore the angle *EDC* equals the angle *BEC.*

But the angle *EBD* is common to the two triangles *BEC* and *BED,* therefore the remaining angle *BED* equals the remaining angle *ECB.* Therefore the triangle *EBD* is equiangular with the triangle *EBC.*

Therefore, proportionally *DB* is to *BE* as *EB* is to *BC.*

But *EB* equals *CD.* Therefore *BD* is to *DC* as *DC* is to *CB.* And *BD* is greater than *DC,* therefore *DC* is also greater than *CB.*

Therefore the straight line *BD* is cut in extreme and mean ratio, and *DC* is its greater segment.

Therefore, *if the side of the hexagon and that of the decagon inscribed in the same circle are added together, then the whole straight line has been cut in extreme and mean ratio, and its greater segment is the side of the hexagon.*

Q.E.D.