L POSTULATES. I. ET it be granted that a ftraight line may be drawn from II. That a terminated straight line may be produced to any length in a ftraight line. III. And that a circle may be defcribed from any centre, at any diftance from that centre. AXIOM S. Book I. T HINGS which are equal to the fame, are equal to one an other. 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 fame, are equal to one another. VII. Things which are halves of the fame, are equal to one another, Magnitudes which coincide with one another, that is, which Straight lines which do not coincide, cannot meet one another in See N. more than one point. XI. All right angles are equal to one another. XII. "If a flraight 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, thefe ftraight lines, being con"tinually produced, fhall at length meet upon that side on "which are the angles which are less than two right angles. "See the notes on Prop. 29. of Book I." PROPO BOOK I. late. T PROPOSITION I. PROBLEM. O defcribe an equilateral triangle upon a given. finite 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 dia3.Poftu flance AB, defcribe a the circle BCD; and from the centre B, at the difiance BA, defcribe the circle ACE; and from the point C, in which the circles cut one bi. 1. Poft. another, draw the ftraight lines b CA, CB to the points A, B; ABC fhall be an equilateral triangle. finition. A B E Because the point A is the centre of the circle BCD, AC is c15th De- equal to AB; and because the point B is the centre of the 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 drft Axi. another d; therefore CA is equal to CB; wherefore CA, AB, BC, are equal to one another; and the triangle ABC is therefore equilateral, and it is described upon the given straight line AB. Which was required to be done. om. a 1. Poft. bl. 1. FRO PROP. II. PROB. ROM a given point to draw a ftraight line equal to a given itraight line. Let A be the given point, and BC the given straight line; it is required to draw from the point A a straight line equal to BC. From the point A to B draw a the ftraight line AB; and upon it defcribeb the equilateral triangle c2. Poft. DAB, and produce the ftraight lines DA, DB, to E and F; from the centre B, at the distance BC, 3. Poft. defcribed the circle CGH; and from the centre D, at the distance DG, defcribe the circle GKL. AL fhall be equal to BC. K H A C B G Because Because the point B is the centre of the circle CGH, BC is Book I. equal e to BG; and becaufe D is the centre of the circle GKL, DL is equal to DG, and DA, DB, parts of them, are equal; e 15. Def. therefore the remainder AL is equal to the remainder f BG: f 3. Ax. 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. PROP. III. PROB. ROM the greater of two given ftraight lines to FR the circle DEF: And because A is the centre of the circle DEF, AE is 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 straight line AE is equal to C, and from AB, c 1. Ax. the greater of two ftraight lines, a part AE has been cut off equal to C the lefs. Which was to be done. IF [F 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; 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 oppofite. 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 BOOK I. AB to DE, and AC to DF; and the angle BAC equal to the angle EDF; the bafe BC fhail 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, shall be equal each to each, viz. the angle ABC to the angle DEF, and the angle ACB to DFE. AA 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, becaufe AB is equal to DE; and AB coinciding with DE, AC shall 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 straight line AC is equal to DF: But the point B coincides with the point E. Therefore the bafe BC fhall coincide with the a 10. Ax. base EF a, and be equal to it b. Wherefore the whole triangle b 8. Ax. ABC fhall coincide with the whole triangle DEF, and be equal to it b; and the other angles of the one shall 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 likewise the angles contained by thofe fides equal to one another, their bases shall 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 PROP. V. THEOR. 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 base Thall be equal. Let ABC be an Ifofceles triangle, of which the fide AB is equal to AC, and let the straight 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. In BD, take any point F, and from AE, the greater, cut off a 3. 1. AG equal a to AF, the less, and join FC, GB. Because D B G b 4. 1. 17 Becaufe AF is equal to AG, and AB to AC, the two fides Book I. 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 b to the base GB, and the triangle AFC to the triangle AGB; and the remaining angles of the one are equal b 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 F angle AGB: And because the whole AF is equal to the whole AG, of which the parts AB, AC are equal; the remainder BF fhall be equal to the c 3. Ax. 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 b, 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 alfo equal; the remaining angle ABC is therefore equal to the remaining angle ACB, which are the angles at the bafe of the triangle ABC: And it has also 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 angles at the bafe, &c. Q. E. D. COROLLARY. Hence every equilateral triangle is alfo equi angular. IF PROP. VI. THEOR. F two angles of a triangle be equal to one another, the fides alfo which fubtend, or are oppofite to, the equal angles, shall be equal to one another. Let ABC be a triangle having the angle ABC equal to the angle ACB; the fide AB is equal to the fide AC. For if AB be not equal to AC, one of them is greater than the other: Let AB be the greater, and from it cut a off DB equal to AC, the lefs, and join DC; therefore, becaufe in the triangles DBC, ACB, DB is equal to AC, and BC common to both, the two fides DB, BC are equal to the two AC, CB, each to each; and the angle DBC is equal to the angle ACB; therefore the base DC is equal to the bafe AB, and the triangle DBC is equal B C to D 23. I |