The second way to execute this probleme.
Suppose the line geuen to be AB. Deuide A•• into two equall parts: as suppose it to be done in the point C. Produce AB from the point B: adioyning a line equall to BC, which let be BD. To the right line AD, and at the point D, erect a perpendicular line equall to BD, let that be DE. Produce ED frō the point D to the point F: making DF to contayne fiue such equall partes, as DE is one. Now vpon EF as a diameter, describe a semicircle which let 〈◊〉〈◊〉 EKF, and let the
point where the circumference of
EKF, doth cut the line
AB, be the point
K. I say that
AB, is deuided in the point
K, by an extreme and meane proportion. For by the 13. of the sixth
ED, DK, &
DF, are three lines in continuall proportion, (
DK being the middle proportionall)
•• Wherefore by the corollary of the 20. of the sixth, as
ED is to
DF, so is the square of
ED, to the square of
DK, but by construction,
ED, is sub∣quintuple to
DF. Wherefore the square of
ED, is subquintuple to the square of
DK. And therefore the square of
DK, is quintuple to the square of
ED. And
ED is equall to
ED, by construction, therefore the square of
DK, is quintuple to the square of
E D. Wherefore the dou∣ble of
BD, is deuided by an extreme and meane proportion: whose greater segment is
BK•• by the second of this thirte
••th. But by con∣struction,
AB, is the double of
••D•• Wherefore
AB, is diuided by ex∣treme and meane proportion, and his greater segment, is
BK: and thereby,
K•• the point of the diuision. We haue therefore deuided by extreme and meane proportion, any right line geuen, in length and po∣sition. Which was to be done.