Stewart's Specific Subjects. EUCLID. [FIRST YEAR.] 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. That a circle may be described from any centre, at any distance from that centre. AXIOMS. one another, I. Things equal to the same are equal to 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. Doubles of the same are equal to one another. VII. Halves or the same are equal to one another. VIII. Magnitudes which coincide with one another, or exactly fill the same space, are equal to one another. IX. The whole is greater than its part. X. Two straight lines cannot enclose a space. PROPOSITION 1.-To describe an equilateral triangle on a given finite straight line. Let AB be the given straight line; it is required to describe an equilateral triangle on it. From the centre A, at the distance AB, describe the circle BCD, and from the centre B, at the distance BA, describe the circle ACE; and from C, where the circles cut one another, join CA, CB; ABC is an equilateral triangle. BE Because A is the centre of BCD, AC is equal to AB; and because B is the centre of 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 same are equal to one another; therefore CA is equal to CB; wherefore CA, AB, BC, are equal to one another and the triangle ABC is equilateral, and is described on the given straight line AB. ; II.—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. Join AB, and on it describe the equilateral triangle DAB, and produce DA, DB, to E and F; from the centre B, at the distance BC, describe the circle CGH, and from the centre D, at the distance DG, describr the circle GKL. Because B is the centre of CGH, BC is equal to BG; and because D is the centre of GKL, DL is equal to DG; and DA, DB, parts of them, are equal; therefore the remainder AL, is equal to the remainder BG; but it has been shown that BC is equal to BG; wherefore AL and BC are each of them equal to BG; and things that K H D A B are equal to the same are equal to one another; therefore 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. III. 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 a part equal to C, the less. From A draw AD equal to C; and EB from the centre A, and at the distance AD, describe the circle DEF; and because A is the centre of DEF, AE shall be equal to AD; but 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 greater of two straight lines, a part AE has been cut off equal to C the less. IV.-If two triangles have two sides of the one equal to two sides of the other, each to each; and likewise the angles contained by those sides equal to one another; they shall likewise have their bases or third sides, equal; and the two 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 sides AB, AC, equal to DE, DF, each to each, viz., AB to DE, and AC to DF; and the angle BAC equal to the angle EDF, the base BC shall be B D CE F equal to the base EF; and the triangle ABC to the |