XXXIV. All other four fided figures besides these, are called Trapeziums. Parallel ftraight lines, are fuch as are in the fame plane, and Book I. LE POSTULATES. I. ET it be granted that a straight line may be drawn from That a terminated straight line may be produced to any length III. And that a circle may be described from any centre, at any distance from that centre. T HINGS which are equal to the fame are equal to one an- II. If equals be added to equals, the wholes are equal. III. If equals be taken from equals, the remainders are equal. IV. If equals be added to unequals, the wholes are unequal. V. If equals be taken from unequals, the remainders are unequal. VI. Things which are double of the same, are equal to one another. Things which are halves of the fame, are equal to one another. Magnitudes which coincide with one another, that is, which IX. "If a ftraight line meets two ftraight lines, fo as to make the "two interior angles on the fame fide of it taken together "lefs than two right angles, these straight 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. T PROBLEM. 15 Book I. OF EUCLID. PROPOSITION I defcribe an equilateral triangle upon a given fi nite ftraight line. Let AB be the given straight line; it is required to describe an equilateral triangle upon it. From the centre A, at the di stance AB, defcribe the circle BCD, and from the centre B, at the distance BA, defcribe the circle ACE; and from the point D A BE 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. 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 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; therefore CA is equal to CB; wherefore CA, AB, d. 1ft Axis BC are equal to one another; and the triangle ABC is there- om. fore equilateral, and it is described upon the given straight line AB. Which was required to be done. PROP. II. PROB. ROM a given point to draw a straight line equal to FR 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 straight 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. Book I. e. 15. Def. 1. 3. Ax. 叉 e 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, DL is equal to DG, and DA, DB, parts of them, are equal; therefore the remainder AL is equal to the remainder f 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 ftraight line BC. Which was to be done. FR PROP. III. PRO B. ROM the greater of two given ftraight lines to cut off a part equal to the less. Let AB and C be the two given 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 a 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 becaufe A is CI. Ax. EB F the center 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 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. PROP. IV. THEOREM. IF 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 likewie have their bases, or third fides, equal; and the two triangles fhail be equal; and their other angles fhall be equal, each to each, viz. those to which the equal fides are oppofite. Le ABC, DEF be two triangles which have the two fides AB, AC equal to the two fides DE, DF, each to each, viz. AB AB to DE, and AC to DF; and the angle BAC equal A to the angle EDF, the base BC fhall be equal to the bafe EF; and the triangle ABC to the triangle DEF; and the other angles, to which the equal fides are oppofite, fhall be equal each to each, viz. the angle ABC to the angle DEF, and the B angle ACB to DFE. СЕ D F Book I. For, if the triangle ABC be applied to DEF, fo that the point A may be on D, and the straight line AB upon DE; the point B fhall coincide with the point E, because AB is equal to DE; and AB coinciding with DE, AC fhall coincide with DF, becaufe 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 ftra ght lines would inclose a space, which is impoffible. Therefore a 10. Axi the base 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 those 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 demonstrated. THE HE angles at the base 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 fhall be equal. Let ABC be an Isofceles triangle, of which the fide AB is e- qual |