1 XXXIV. All other four sided figures besides these, are called Trapeziums. Xxxv. Parallel straight lines, are such as are in the fame plane, and L ET POSTULATES. I. it be granted that a straight line may be drawn from any one point to any other point. II. That a terminated straight line may be produced to any length in a straight line. III. And that a circle may be described from any centre, at any ΑΧΙOMS. I. HINGS which are equal to the fame are equal to one an Tother. 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. VII. Things which are halves of the fame, are equal to one another. VIII. Magnitudes which coincide with one another, that is, which * Book I. "If a straight line meets two straight lines, so as to make the " 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 fide on " which are the angles which are less than two right angles, "See the notes on Prop. 29. of Book I." 4 1 PROPO. OF EUCLID. PROPOSITION I. PROBLEM. O describe an equilateral triangle upon a given fi- Let AB be the given straight line; it is required to describe an equilateral triangle upon it. 15 Book I. From the centre A, at the di C stance AB, describe the circle BCD, and from the centre B, at the distance BA, describe the a. 3. Postulate. circle ACE; and from the point D A BE C, in which the circles cut one another, draw the straight lines CA, CB to the points A, B; ABC shall be an equilateral triangle. b b. 2d. Poft. ! 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 5th 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. ist 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. F PROP. II. PROB. ROM a given point to draw a straight line equal to Let A be the given point, and BC the given straight line; it is From the point A to B draw the straight line AB; and upon it describe the equilateral triangle DAB, and produce the straight lines DA, DB, to E and F; from the centre B, at the distance ВС, described the circle CGH, and from the centre D, at the distance DG, describe the circle GKL. AL Thall be equal to BC, Book I. 8. 15. Def. 1. 3. Ax. 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 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 fame 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. ROM the greater of two given straight lines to cut a part equal to the less. FRO Let AB and C be the two given straight lines, whereof AB is the greater. It is required to cut off from AB, the greater, a part equal to C the less. D C 8.2 1. C. 1. Ax. From the point A draw a the straight line AD equal to C; and from the centre A, and at b. 3. Post. the distance AD, defcribe the 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 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 EB F 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 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 the equal fides are opposite. 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; angle ACB to DFE. 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 shall 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 bafe EF, because the point B coinciding with E, and C with F, if the base BC does not coincide with the base EF, two straght lines would inclose a space, which is impoffible. 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 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 those 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 opposite shall be equal, each to each. Which was to be demonstrated. a 10. Ax HE angles THE at the base of an Isosceles triangle are equal to one another; and, if the equal fides be produced, the angles upon the other fide of the base shall be equal. Let ABC be an Isosceles triangle, of which the side AB is e B qual : |