The elements of geometrie of the most auncient philosopher Euclide of Megara. Faithfully (now first) translated into the Englishe toung, by H. Billingsley, citizen of London. Whereunto are annexed certaine scholies, annotations, and inuentions, of the best mathematiciens, both of time past, and in this our age. With a very fruitfull præface made by M. I. Dee, specifying the chiefe mathematicall scie[n]ces, what they are, and wherunto commodious: where, also, are disclosed certaine new secrets mathematicall and mechanicall, vntill these our daies, greatly missed

The 8. Probleme. The 8. Proposition. If in an equilater & equiangle Pētagon two right lines do subtend two of the angles following in order: those lines doo diuide the one the other by an extreme and meane proportion: and the greater segments of those lines are ech equall to the side of the Pentagon.

SVppose that ABCDE be an equilater and equiangle Pentagon. And let two right lines AC, and BE, subtend the two angles A, and B, which follow in order. And let them cut the one the other in the point H. Then I say that either of those lines is diuided by an extreme & meane proportiō in the point H: And that eche of the greater segments of those lines are equal to the side of the Pentagon Circumscribe (by the 14. of the fourth) about the Pentagō. ABCDE,* 1.1 a circle ABCDE. And forasmuch as these two right lines EA, and AB, are equall to these two right lines AB, and BC,* 1.2 and they contayne equall angles: therefore (by the 4. of

[illustration]

the first) the base BE, is equal to the base AC: and the triangle ABE, is equall to the triangle ABC, and the angles remayning, are equall to the angles remay∣nyng, the one to the other, vnder which are sub∣tended equall sides. Wherefore the angle BAC, is equall to the angle ABE. Wherfore the angle AHE is double to the angle BAH, (by the 32. of the first) for it is an outward angle of the triangle ABH. And the angle EAC is double to the angle•• BAC (by the last of the sixth). For the circumference EDC is dou∣ble to the circumference CB. Wherefore the angle H∣AE is equall to the angle AHE. Wherefore also the right line ••E, is (by the 6. of the first) equall to the right line EA, that is to the line AB. And forasmuch as the right line BA is equall to the right line AE, therefore the angle ABE is equall to the angle AEB. But it is proued that the angle ABE is equal to the angle BAH: wherefore also the angle BEA is equall to the angle BAH. And in the two triangles ABE, and ABH, the angle ABE is common to them both, wherefore the angle remayning, namely, BAE is equall to the angle remayning, namely, to AHB (by the corollary of the 32. of the first). Wherfore the triangle ABE, is e∣quiangle to the triangle ABH. Wherefore proportionally as the line EB, is to the line BA, so is the line AB to the line BH (by the 4. of the sixth). But the line BA is equall to the line EH. Wherefore as the line B•• is to the line EH, so is the line EH to the line H••. But the line BE is greater the•• the line BA•• wherefore the line EH, also is greater then the line H∣B. Wherefore the line BE•• is diuided by an extreme and meane proportion in the point H (by the 3. diffinition of the sixth) and his greater segment EH is equall to the side of the Penta∣gon. In like sort also may we proue that the line AC is diuided by an extreme and meane pro∣portion in the point H, and that his greater segment CH, is equall to the side of the Penta∣gon. (For the whole line AC is equall to the whole line BE, and it hath bene proued that the parts taken away BH, and AH are equall wherfore the residue CH is equall to the residue EH (by the 10. of the fifth). If therefore in an equilater and equiangle Pentagon two right lines 〈◊〉〈◊〉 sub••end two of the angles following in order: those lines doo diuide the one the o∣ther by an extreme and meane proportion: and the greater segments of those lines are eche equal to the side of the Pentagon: which was required to be demonstrated.

Notes

• * 1.1

  Construction.
• * 1.2

  Demonstra∣tion.