"If a ftraight line meets two straight lines, fo as to make the "two interior angles on the fame fide of it taken together "less than two right angles, thefe ftraight lines being con"tinually produced, fhall at length meet upon that fide on "which are the angles which are lefs than two right angles. "See the notes on Prop. 29. of Book I.” PROPO. PROPOSITION I PROBLEM. O defcribe an equilateral triangle upon a given fi- Let AB be the given ftraight line; it is required to defcribe an equilateral triangle upon i it. From the centre A, at the di-, ftance AB, defcribe the circle BCD, and from the centre B, at. the distance BA, defcribe the circle ACE; and from the point D C, in which the circles cut one another, draw the ftraight lines b CA, CB to the points A, B; ABC fhall be an equilateral triangle. Book I. a. 3. Poftulate. finition. Because the point A is the centre of the circle BCD, AC is equal to AB; and because the point B is the centre of the c. 15th De circle ACE, BC is equal to BA: But it has been proved that CA is equal to AB; therefore CA, CB are each of them equal to AB; but things which are equal to the fame are equal to one another d; therefore CA is equal to CB; wherefore CA, AB, d. 1ft AxiBC are equal to one another; and the triangle ABC is there-om. fore equilateral, and it is defcribed upon the given ftraight line AB. Which was required to be done, PROP. II. PROB. ROM a given point to draw a ftraight line equal to a given straight line. FROM Let A be the given point, and BC the given ftraight line; it is required to draw from the point A a ftraight line equal to BC. From the point A to B draw a the ftraight line AB; and upon it defcribe the equilateral triangle DAB, and produce the_ftraight lines DA, DB, to E and F; from the centre B, at the distance BC, defcribed the circle CGH, and from the centre D, at the distance DG, defcribe the circle GKL. AL hall be equal to BC. D K H B d. 3. Poft. G E Because Book I. f Because the point B is the centre of the circle CGH, BC is equal to BG; and because D is the centre of the circle GKL, e. 15. Def. DL is equal to DG, and DA, DB, parts of them, are equal; f. 3. Ax. therefore the remainder AL is equal to the remainder £ BG: But it has been fhewn, that BC is equal to BG; wherefore AL and BC are each of them equal to BG; and things that are equal to the fame are equal to one another; therefore the ftraight line AL is equal to BC. Wherefore from the given point A a ftraight line AL has been drawn equal to the given Atraight line BC. Which was to be done. a. 2. I. F 1 PROP. III. PROB. ROM the greater of two given ftraight lines to cut off a part equal to the less. Let AB and C be the two gi ven ftraight lines, whereof AB is the greater. It is required to cut off from AB, the greater, a part equal to C the lefs. From the point A draw the ftraight line AD equal to C; and from the centre A, and at b. 3. Poft. the distance AD, defcribe the circle DEF; and because A is EB the centre of the circle DEF, AE fhall be equal to AD; but the ftraight line C is likewife equal to AD; whence AE and C are each of them equal to AD; wherefore the ftraight line AE is c. I. Ax. equal to C, and from AB, the greater of two ftraight lines, a part AE has been cut off equal to C the lefs. Which was to be done. с PROP. IV. THEOREM. two triangles have two fides of the one equal to two fides of the other, each to each; and have likewise the angles contained by thofe fides equal to one another; they fhall likewife have their bafes, or third fides, equal; and the two triangles fhall be equal; and their other angles fhall be equal, each to each, viz. thofe to which the equal fides are opposite. Let ABC, DEF be two triangles which have the two fides AB, AC equal to the two fides DE, DF, each to each, viz. AB For, if the triangle ABC be applied to DEF, fo that the point. A may be on D, and the ftraight line AB upon DE; the point. B fhall coincide with the point E, becaufe AB is equal to DE; and AB coinciding with DE, AC fhall coincide with DF, because the angle BAC is equal to the angle EDF; wherefore alfo the point C fhall coincide with the point F, because the ftraight line AC is equal to DF: But the point B coincides with the point E; wherefore the base BC fhall coincide with the base EF, because the point B coinciding with E, and C with F, if the bafe BC does not coincide with the bafe EF, two ftraight lines would inclofe a fpace, which is impoffible. Therefore a 10. Ax. the bafe BC fhall coincide with the bafe EF, and be equal to it. Wherefore the whole triangle ABC fhall coincide with the whole triangle DEF, and be equal to it; and the other angles of the one fhall coincide with the remaining angles of the other, and be equal to them, viz. the angle ABC to the angle DEF, and the angle ACB to DFE. Therefore, if two triangles have two fides of the one equal to two fides of the other, each to each, and have likewife the angles contained by thofe fides equal to one another, their bafes fhall likewife be equal, and the triangles be equal, and their other angles to which the equal fides are oppofite fhall be equal, each to each. Which was to be demonftrated. TH 'HE angles at the bafe of an Ifofceles triangle are equal to one another; and, if the equal fides be produced, the angles upon the other fide of the bafe hall be equal. Let ABC be an Ifofceles triangle, of B h the fide AB is e qual Book I. qual to AC, and let the ftraight lines AB, AC be produced to D and E, the angle ABC fhall be equal to the angle ACB, and the angle CBD to the angle BCE. a 3. 1. b 4. I. 3. Ax. In BD take any point F, and from AE, the greater, cut off AG equal to AF, the lefs, and join FC, GB. A Because AF is equal to AG, and AB to AC, the two fides FA, AC are equal to the two GA, AB, each to each; and they contain the angle FAG common to the two triangles AFC, AGB; therefore the bafe FC is equal to the bafe GB, and the triangle AFC to the triangle AGB; and the remaining angles of the one are equal to the remaining angles of the other, each to each, to which the equal fides are oppofite; viz. the angle ACF to the angle ABG, and the angle AFC to the angle AGB: And because the whole AF is equal to the whole AG, of which the parts AB, AC, are equal; the re F B с G E mainder BF fhall be equal to the remainder CG; and FC was proved to be equal to GB; therefore the two fides BF, FC are equal to the two CG, GB, each to each; and the angle BFC is equal to the angle CGB, and the bafe BC is common to the two triangles BFC, CGB; wherefore the triangles are equal', and their remaining angles, each to each, to which the equal fides are oppofite; therefore the angle FBC is equal to the angle GCB, and the angle BCF to the angle CBG: And, fince it has been demonftrated, that the whole angle ABG is equal to the whole ACF, the parts of which, the angles CBG, BCF are allo equal; the remaining angle ABC is therefore equal to the re maining angle ACB, which are the angles at the base of the triangle ABC: And it has alfo been proved that the angle FBC is equal to the angle GCB, which are the angles upon the other fide of the bafe. Therefore the angies at the bafe, &c. Q E. D. COROLLARY. Hence every equilateral triangle is alfo equiangular. F two angles of a triangle be equal to one another, the fides alfo which fubtend, or are oppofite to, the equal angles, fhall be equal to one another. Let |