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 11. Probleme. The 42. proposition. Vnto a triangle geuen, to make a parallelograme equal, whose angle shall be equall to a rectiline angle geuen.

SVppose that the triangle geuen be ABC, and let the rectiline angle geuen be D. It is required that vnto the triangle ABC there be made a parallelograme equall, whose angle shal be equall to the rectiline angle geuen, namely,* 1.1 to the angle D. Deuide (by the 10. propositiō) the line BC into two equall partes in the pointe E. And (by the first peticion) draw a right line from the point A to the point E. And (by the 23. proposition) vpon the right line geuen EC, and to the point in it

[illustration]

geuen E, make the angle CEF equal to the angle D. And (by the 31. pro∣position) by the point A draw vnto the line EC a parallel line AH: and let the line EF cut the line AH in the point F. and (by the same) by the point C, drawe vnto the line EF a parallel line CG. VVherfore FECG is a parallelograme.* 1.2 And forasmuche as BE is equall to EC, therfore (by the 38. proposition) the triangle ABE is equall to the triangle AEC, for they consist vpō equall bases that is BE and EC, and are in the selfe same parallel lines, namely, BC and AH. VVherfore the triangle ABC is double to the triangle AEC. And the paralle∣lograme CEFG is also double to the triangle AEC: for they haue one & the selfe same base, namely, EC: and are in the selfe same parallel lines, that is, EC and AH. VVherfore the parallelograme FECG is equall to the triangle ABC, and hath the angle CEF equall to the angle geuen D. VVherefore vnto the triangle geuen ABC is made an equall parallelograme, namely, FECG, whose angle CEF is equall to the angle geuen D: which was required to be done.

The conuerse of this proposition after Pelitarius.

Vnto a parallelogramme geuen,* 1.3 to make a triangle equall, hauy••g an angle equall to a rectiline angle geuen.

Suppose that the parallelograme geuen be ABCD, and let the angle geuen be E. It is required vnto the parallelograme ABCD to make a triangle equall hauyng an

angle equal to the angle E. Vpon

[illustration]

the line CD and to the pointe in it C, describe (by the 23. propo∣position) an angle equall to the angle E: which let be DCF: a••d let the line CF cut the line AB being produced, in the point F: and produce the line CD (which is parallell to the line AF) to the point G. And put the line DG e∣quall to the line CD and draw a line from F to G. Then I say that the triangle CFG is equal to the parallelograme ABCD. For for∣asmuch as (by the 38. proposition) the whole triangle CFG is double to the triangle CDF. A••d (by the 41. proposition) the parallelograme ABCD is double to the same triangle CDF: therfore the parallelograme ABCD and the triangle CFG are equall the one to the other: which was required to be done.

Notes

• * 1.1

  Construction.
• * 1.2

  Demonstration
• * 1.3

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