See N. XXIX. An acute-angled triangle is that which has XXX. Of four-sided figures, a square is that which XXXI. An oblong is that which has all its angles XXXII. A rhombus is that which has all its XXXIII. A rhomboid is that which has its opposite 0 XXXIV. All other four-sided figures besides these, are called XXXV. Parallel straight lines are such as are in 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. 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 same, 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. Two straight lines cannot enclose a space. XII. "If a straight line meet two straight lines, so as to "make the two interior angles on the same side of it taken "together, less than two right angles, these straight lines. "being continually produced, shall at length meet upon "that side on which are the angles which are less than two " right angles." See the notes on Prop. XXIX. of Book I. • 3 Postulate. • 1 Post. 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 AB. 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 the point 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. BE 15 Definition. * 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 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 same thing are * 1 Axiom. equal to one another; 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. * * I Post. * 1. 1. • 2 Post. 3 Post. PROPOSITION 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 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 BC, describe the circle CGH, and from the cen tre D, at the distance DG, describe the circle GKL: AL shall be equal to BC. 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 BG: but it has been 15 Def. + Constr. 3 Ax. 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 thing are equal to one another; therefore the straight line † 1 Ax. 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. PROPOSITION 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. 2. 1. From the point A, draw the straight line AD equal to C; and from the centre A, and at the distance AD, describe the circle DEF: AE shall be equal to C. Α EB 3 Post. F Because A is the centre of the circle DEF, AE is equal to † 15 Def. AD; but the straight line C is likewise equal to† AD; whence † Constr. AE and C are each of them equal to AD; wherefore, the straight line AE is equal to two straight lines, a part AE less. Which was to be done. C, and from AB, the greater of Ax. has been cut off, equal to C the PROPOSITION 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, 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 cach, viz. those to which the equal sides are opposite. + Hyp. ↑ Hyp. † Hyp. * 10 Ax. ↑ 8 Ax. D 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 DE, and AC to DF; and the angle BAC equal to the angle EDF: the base BC shall be equal to the base EF; and the triangle ABC to the triangle DEF; and the other angles, to which the equal B sides are opposite, shall be equal, each to Да CE each, viz. the angle ABC to the angle DEF, and the 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 was proved to coincide 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 straight lines would inclose a space, which is impossible: therefore the base BC coincides with the base EF, and therefore is equal to it. Wherefore the whole triangle ABC coincides with the whole triangle DEF, and is equal to it; and the other angles of the one, coincide with the remaining angles of the other, and are 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 those sides equal to one another, their bases shall likewise be equal; and the triangles shall be equal, and their other angles, to which the equal sides are opposite, shall be equal, each to each. Which was to be de monstrated. PROPOSITION V. PONS A SYNORUM. THEOR.-The 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 shall be equal. Let ABC be an isosceles triangle, of which the said AB is equal to AC; and let the straight lines AB, AC be produced |