33. A rhomboid is that which has its opposite sides equal to one another, but all its sides are not equal, nor its angles right angles : 34. All other four-sided figures besides these are called trapeziums. 35. Parallel straight lines are such as are in the same plane, and which being produced ever so far both ways do not meet. [Note. The terms oblong and rhomboid are not often used. Practically the following definitions are used. Any four-sided figure is called a quadrilateral. A line joining two opposite angles of a quadrilateral is called a diagonal. A quadrilateral which has its opposite sides parallel is called a parallelogram. The words square and rhombus are used in the sense defined by Euclid; and the word rectangle is used instead of the word oblong. Some writers propose to restrict the word trapezium to a quadrilateral which has two of its sides parallel ; and it would certainly be convenient if this restriction were universally adopted.] POSTULATES. Let it be granted, 1. That a straight line may be drawn from any one point to any other point: 2. That a terminated straight line may be produced to any length in a straight line : 3. And that a circle may be described from any centre, at any distance from that centre. AXIOMS. 1. Things which are equal to the same thing are equal to one another. 2. If equals be added to equals the wholes are equal. 3. If equals be taken from equals the remainders are equal. 4. If equals be added to unequals the wholes are unequal. 5. If equals be taken from unequals the remainders are unequal. 6. Things which are double of the same thing are equal to one another. 7. Things which are halves of the same thing are equal to one another. 8. Magnitudes which coincide with one another, that is, which exactly fill the same space, are equal to one another. 9. The whole is greater than its part. 12. 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 on that side on which are the angles which are less than two right angles. PROPOSITION 1. PROBLEM. 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 AB. From the centre A, at the distance AB, describe the circle BCD. [Postulate 3. From the centre B, at the distance BA, describe the circle ACE. [Postulate 3. From the point C, at which the circles cut one another, draw the straight lines CA and CB to the points A and B. [Post. 1. ABC shall be an equilateral triangle. Because the point A is the centre of the circle BCD, AC is equal to AB. [Definition 15. And because the point B is the centre of the circle ACE, BC is equal to BĂ. [Definition 15. But it has been shewn that CA is equal to AB; therefore CA and CB are each of them equal to AB. But things which are equal to the same thing are equal to one another. [Axiom 1. Therefore CA is equal to CB. Therefore CA, AB, BC are equal to one another. Wherefore the triangle ABC is equilateral, [Def. 24. and it is described on the given straight line AB. Q.E.F. PROPOSITION 2. PROBLEM. From a given point to draw a straight line equal to a giden 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; [Post. 1. and on it describe the equilateral triangle DAB, [I. 1. H and produce the straight lines DA, DB to E and F. [Post. 2. ВІ G F Because the point B is the centre of the circle CGH, BC is equal to BG. [Definition 15. And because the point D is the centre of the circle GKL, DL is equal to DG; [Definition 15. and DA, DB parts of them are equal ; [Definition 24. therefore the remainder AL is equal to the remainder BG. [Axiom 3. But it has been shewn that BC is equal to BG; therefore AL and BC are each of them equal to BG. But things which are equal to the same thing are equal to one another. [Axiom 1. 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. Q.E.F. PROPOSITION 3. PROBLEM. From the greater of two given straight lines to cut of a part equal to the less Let AB and C be the two given straight lines, of which 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 the straight line AD equal to C; [I. 2. and from the centre A, at the distance AD, describe B the circle DEF meeting AB at E. [Postulate 3. AE shall be equal to C. Because the point A is the centre of the circle DEF, AE is equal to AD. [Definition 15. But C is equal to AD. [Construction. Therefore AE and C are each of them equal to AD. Therefore AE is equal to C. [Axiom 1. Wherefore from AB the greater of two given straight lines a part AE has been cut off equal to the less. Q.E.F. PROPOSITION 4. THEOREM. If two triangles have two sides of the one equal to two sides of the other, each to each, and have also the angles contained by those sides equal to one another, they shall also have their bases or third sides equal; and the two triangles shall be equal, and their other angles shall be equal, each to each, namely 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, namely, 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 shall be equal, each B to each, to which the equal sides are opposite, namely, the angle ABC to the angle DEF, and the angle ACB to the angle DFE. ДА |