XXX. Book I. Straight lines, which are in the same plane, and being produced ever so far both ways, do not meet, are called Parallel Lines. N. POSTULATES. I. LET 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 distance from that centre. AXIOMS. I. THINGS which are equal to the same thing are equal to one another. 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. 1 VI. Things which are doubles of the same thing, are equal to *Book I. one another. VII. Things which are halves of the same thing are equal to one another. VIII. Magnitudes which coincide with one another, that is, which exactly fill the same space, are equal to one another. IX. The whole is greater than its part. X. All right angles are equal to one another. XI. "Two straight lines which intersect one another, cannot "be both parallel to the same straight line.” Book I. late. PROPOSITION I. PROBLEM. To describe an equilateral triangle upon a given finite straight line. Let AB be the given straight line; it is required to describe an equilateral triangle upon From the centre A, a 3. Postu- scribe the circle BCD, b. 1. Post. straight lines b CA, nition. triangle. it. Because the point A is the centre of the circle BCD, c. 11. Defi- AC is equal to AB; and because the point B is the centre of the circle ACE, BC is equal to AB: But it has been proved that CA is equal to AB; therefore CA, CB are each of them equal to AB; now things which are d 1. Axiom. equal to the same are equal to one another; therefore CA is equal to CB; wherefore CA, AB, CB are equal to one another; and the triangle ABC is therefore equilateral, and it is described upon the given straight line Which was required to be done. a 1. Post. b. 1. 1. AB. PROP. II. PROB. From a given point to draw a straight line equal to a given straight 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 straight line AB; and upon it describe the equilateral triangle DAB, and produce the straight K lines DA, BD, to E and F; Book I. c. 2. Post. centre D, at the distance from the centre B, at the H distance BC, described the d 3. Post. circle CGH, and from the DG, describe the circle GKL. C Because the point B is the centre of the circle CGH, of circle GKL, DL is equal to DG, and DA, DB, parts them, are equal; therefore the remainder AL is equal to the remainder BG: But it has been shewn, that BC f3. Ax. 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. PROP. III. PROB. From the greater of two given straight lines to cut off a part equal to the less. 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. From the point A draw a the straight line AD equal to C; and from the centre A, and at the dis b E B a 2. 1. tance AD, describe the circle DEF; and because Ab 3. Post. is the centre of the circle DEF, AE is equal to AD; but the straight line C is likewise 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 the c 1. Ax. Book I. greater of two straight lines, a part AE has been cut off equal to C the less. Which was to be done. N. PROP. IV. THEOREM. 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 those sides equal to one another, their bases or third sides, shall be equal; and the areas of the triangles shall be equal; and their other angles shall be equal, each to each, viz. those to which the equal sides are opposite*. Let ABC, DEF be two triangles which have the two sides AB, AC equal to the two sides DE, DF, each to each, viz. AB to A triangle DEF; B CE D and the other angles, to which the equal sides are opposite, shall be equal, each to each, viz. the angle, ABC to the angle DEF, and the angle ACB to DFE. For, if the triangle ABC be applied to the triangle 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 AC is equal to DF: But the point B coincides with the point E; wherefore the base BC shall coincide with the a cor. def. 3. base EF a, and shall be equal to it. Therefore also the whole triangle ABC shall coincide with the whole tri * The three conclusions in this enunciation are more briefly expressed by saying, that the triangles are every way equal. |