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 30. Theoreme. The 35. Proposition. If there be two superficiall angles equall, and from the pointes of those an∣gles

be eleuated on high right lines, comprehending together with those right lines which containe the superficiall angles, equall angles, eche to his corespōdent angle, and if in eche of the eleuated lines be takē a point at all auentures, and from those pointes be drawen perpendicular lines to the ground playne superficieces in which are the angles geuen at the begin∣ning, and from the pointes which are by those perpendicular lines made in the two playne superficieces be ioyned to those angles which were put at the beginning right lines: those right lines together with the lines eleua∣ted on high shall contayne equall angles.

SVppose that these two rectiline superficiall angles BAC, and EDF be equall the one to the other:* 1.1 and from the pointes A and D let there be eleuated vpward these right lines AG and DM, comprehendinge together with the lines put at the beginninge equall angles, ech to his correspondent angle, that is, the angle MDE to the angle GAB, and the angle MDF to the angle GAC, and take in the lines AG and DM pointes at all auētures and let the same be G and M. And (by the 11. of the eleuēth) from the pointes G and M draw vnto the ground playne superficieces wherein are the an∣gles BAC and E

[illustration]

DF perpendicular lines GL and MN and let them fall in the sayd playne super••icieces in the pointes N and L, and drawe a right line from the point L to the point A and an other from the pointe N to the pointe D. Then I say that the angle GAL is equall to the angle MDN. Frō the greater of the two lines AG and DM, (which let be AG) cut of by the 3. of the first the line AH equall vnto the line DM. And (by the 31. of the first) by the point H, drawe vnto the line GL a parallel line, and let the same be HK. Now the line GL is erected perpendicularly to the grounde playne superficies BAL: Wherfore also (by the 8. of the eleuenth) the line HK is erected perpēdicularly to the same grounde plaine superficies BAC. Drawe (by the 12. of the first) frō the pointes K and N vnto the right lines AB, AC, DF, & DE perpēdicular right lines, and let the same be KC, NF, KB, NE. And drawe these right lines HC, CB, MF, FE. Now forasmuch a•• (by the 47. of the first) the square of the line HA is equall to the squares of the lines HK and KA,* 1.2 but vnto the square of the line KA are equall the squares of the lines KC and CA: Wherefore the square of the line HA is equall to the squares of the lines HK, KC and CA. But by the same vnto the squares of the lines HK and KC is equall the square of the line HC: Wherefore the square of the line HA is equall to the squares of the lines HC and CA: wherfore the angle HCA is (by the 48. of the first) a right angle. And by the same reason also the angle MFD is a right angle. Wherefore the angle HCA is equall to the angle MFD. But the angle HAC is (by suppositiō) equal to the angle MDF. Wherfore there are two triangles MDF and HAC hauing two an∣gles of the one equall to twoo angles of the other, eche to his correspondent angle, and one side of the one equall to one side of the other, namely, that side which subtendeth one of the equall

angles, that is, the side HA is equall to the side DM by construction. Wherefore the sides remayning are (by the 26. of the first) equall to the sides remayning. Wherefore the side AC is equall to the side DF. In like sort may we proue that the side AB is equall to the side DE, if ye drawe a right line from the point H to the point B, and an other from the point M to the point E. For forasmuch as the square of the line AH is (by the 47. of the firste) equall to the squares of the lines AK and KH, and (by the same) vnto the square of the line AK are equall the squares of the lines AB and BK. Wherefore the squares of the lines AB, BK, and KH are equall to the square of the line AH. But vnto the squares of the lines BK and KH is equall the square of the line BH (by the 47. of the first) for the angle HKB is a right angle, for that the line HK is erected perpēdicularly to the ground playne superficies: Wherefore the square of the line AH is equall to the squares of the lines AB and BH. Wherefore (by the 48. of the first) the angle ABH is a right angle. And by the same rea∣son the angle DEM is a right angle. Now the angle BAH is equall to the angle EDM, for it is so supposed, and the line AH is equall to the line DM. Wherefore (by the 26. of the firste) the line AB is equall to the line DE. Now forasmuch as the line AC is equall to the line DF, and the line AB to the line DE, therefore these two lines AC and AB are equall to these two lines FD and DE. But the angle also CAB is by suppositi∣on equall to the angle FDE. Wherefore (by the 4. of the firste) the base BC is equall to the base EF, and the triangle to the triangle, and the rest of the angles to the reste of the angles. Wherefore the angle ACB is equall to the angle DFE. And the right angle ACK is equal to the right angle DFN. Wherfore the angle remayning, namely, BCK, is equall to the an∣gle remayning, namely, to EFN. And by the same reasō also the angle CBK is equal to the angle FEN. Wher¦fore

[illustration]

there are two triangles BCK, & EFN, hauing two angles of the one equal to two angles of the other, eche to his correspondent angle, and one side of the one equall to one side of the o∣ther, namely, that side that lieth betwene the equall angles, that is the side BC is equall to the side EF: Where∣fore (by the 26. of the first) the sides remaininge are equall to the sides remayning. Wherfore the side CK is equall to the side FN: but the side AC is equall to the side DF. Wherefore these two sides AC and CK are equall to these two sides DF and FN, and they contayne equall angles: Wherefore (by the 4. of the first) the base AK is equall to the base DN. And forasmuch as the line AH is equall to the line DM, therefore the square of the line AH is equall to the square of the line DM. But vnto the square of the line AH are equall the squares of the lines AK and KH (by the 47. of the first) for the angle AKH is a right angle. And to the square of the line DM are equall the squares of the lines DN and NM, for the angle DNM is a right angle. Wherefore the squares of the lines AK and KH are equall to the squares of the lines DN and NM: of which two, the square of the line AK is equall to the square of the line DN (for the line AK is proued equall to the line AN). Wherefore the residue, namely, the square of the line KH is equal to the residue, namely, to the square of the line NM. Wherefore the line HK is equall to the line MN. And forasmuch as these two lines HA and AK are equall to these two lines MD and DN, the one to the other, and the

base HK is equall to the base MN: therfore (by the 8. of the first) the angle HAK is equall to the angle MDN. If therefore there be two superficiall angles equall, and frō the pointes of those angles be eleuated on high right lines, comprehending together with those right lines which were put at the beginning, equall angles, ech to his corespondent angle, and if in ech of the erected lines be taken a point at all aduentures, and from those pointes be drawen perpen∣dicular lines to the plaine superficieces in which are the angles geuen at the beginning, and fr•••• the pointes which are by the perpendicular lines made in the two plaine superficieces be ioyned right lines to those angles which were put at the beginning, those right lines shall to∣gether with the lines eleuated on high make equall angles which was required to be proued.

Because the figures of the former demonstration are somewhat hard to conceaue as they are there drawen in a plaine, by reason of the lines that are imagined to be eleuated on high, I haue here set o∣ther figures, wherein you must e∣recte

[illustration]

perpendicularly to the ground superficieces the two triangles BHK, and EMN, and then ele∣uate the triangles DFM, & ACH, in such sort that the angles M and H of these triangles, may concurre with the angles M and H of the o∣ther erected triangles. And then imagining only a line to be drawen from the point G of the line AG to the point L in the ground superfi∣cies, compare it with the former construction & demonstration, and it will make it very easye to con∣ceaue.

¶ Corollary.

By this it is manifest, that if there be two rectiline superficiall angles e∣quall, and vpon those angles be eleuated on high equall right lines contay∣ning together with the right lines put at the b••ginning equall angles: per∣pendicular lines drawen from those eleuated lines to the ground plaine su∣perficieces wherein are the angles put at the beginning, are equall the one to the other. For it is manifest, that the perpendicular lines HK, & MN, which are dra∣wen from the endes of the equall eleuated lines AH, and DM, to the ground superficieces, are equall.

Notes

• * 1.1

  Construction.
• * 1.2

  Demonstra∣tion.