Book I. XXXIV. XXXV. which, being produced ever so far both ways, do not meet. POS T U L A T E S. 1 L I, II. III. distance from that centre. T A X I 0 MS. I. II. III. IV. V. VI. VII. VIII. exactly fill the same space, are equal to one another. IX. Book 1. IX. x. XI. XII. “two interior angles on the same fide of it taken together « less than two right angles, these straight lines being con“ tinually produced, shall at length meet upon that side on s which are the angles which are less than two right angles, " See the notes on Prop. 29. of Book I.”. PROPO. Tn a late. PROPOSITION I. PROBLEM. Book I. 10 describe an equilateral triangle upon a given fi nite straight line. Let AB be the given straight line; it is required to describe an equilateral triangle upon it. From the centre A, at the distance AB, describe the circle a. 3. PoftuBCD, and from the centre B, at the distance BA, describe the circle ACE ; and from the point D B E C, in which the circles cut one Isi another, draw the straight lines CA, CB to the points A, B; ABC b.zd. Poft, Thall be an equilateral triangle. 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!sth De circle ACE, BC is equal to BA: But it has been proved that CA finition PROP. II. PROB. a given Itraight line. F b. I.I. 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 K a. 1. Poft. the straight line AB ; and upon it describe b the equilateral triangle H DAB, and produce the straight c. 2. Poft. lines DA, DB, to E and F; from D the centre B, at the distance BC, described the circle CGH, and d. 3. Post from the centre D, at the distance E DG, describe the circle GKL. AL fhall be equal to BC. F] Becautc Book I. e F Because the point B is the centre of the circle CGH, BC is equal e 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; , 3. Ax. therefore the remainder AL is equal to the remainder f BG: But it has been shewn, that BC is equal to BG; wherefore AL and BC are each of them equal to BG; and things that are equal to the same are equal to one another; therefore the straight line AL is equal to BC. Wherefore from the given point A a straight line AL has been drawn equal to the given straight line BC. Which was to be done. PRO P. III. PROB. off a part equal to the less E B and from the centre A, and at b. 3. Poft. the distance AD, describe o the F circle DEF; and because A is the center of the circle DEF, AE shall be equal to AD; but the straight line C is likewise equal to AD; whence AE and Care each of them equal to AD, wherefore the straight line AE is 6 1. Ax. equal to C, and from AB, the greater of two straight lines, a part AE has been cut off equal to C the less. Which was to PROP. IV. THEOREM. fides of the other, each to each; and have likewise the angles contained by those fides equal to one another; they shall likewise have their bases, or third fides, equal ; and the two triangles shall be equal ; and their other angles shall be equal, each to each, viz. those to which thc equal sides are opposite. Le ABC, DEF be two triangles which have the two fides) AB, AC equal to the two sides DE, DF, each to each, viz. AB с Book 1. { AB to DE, and AC to DF; D For, if the triangle ABC be applied to DEF, so that the point A may be on D, and the straight line AB upon DE ; the point B shall coincide with the point E, because 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 also the point C shall coincide with the point F, because the straight line AC is equal to DF: But the point B coincides with the point E; wherefore the base BC shall coincide with the base EF, because the point B coinciding with E, and C with F, if the base BC does not coincide with the base EF, two tra ght lines would inclose a space, which is impossible a. Therefore the base BC shall coincide with the base EF, and be equal to it. Wherefore the whole triangle ABC shall coincide with the whole triangle DEF, and be equal to it; and the other angles of the one thall 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 sides of the one equal to two sides of the other, each to each, and have likewise the angles contained by chose fides equal to one another, their bases thall likewife be equal, and the triangles be equal, and their other angles to which the equal fides are opposite shall be equal, each to each. Which was to be demonstrated. a 10. Ax: TH 'HE angles at the base of an Isosceles triangle are equal to one another; and, if the equal sides be produced, the angles upon the other side of the base İhall be equal. Let ABC be an Isosceles triangle, of which the fide AB is e. B qual |