27. A right-angled triangle is that which has a right angle. 28. An obtuse-angled triangle is that which has an obtuse angle. 29. An acute-angled triangle is that which has three acute angles. 30. Of four-sided figures, a square is that which has all its sides equal, and all its angles right angles. 31. An oblong is that which has all its angles right angles, but not all its sides equal. 32. A rhombus is that which has all its sides equal, but its angles are not right angles. 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. Any four-sided figure is called a Quadrilateral. 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 1. A Parallelogram is a quadrilateral which has its oppo site sides parallel. 2. A Rectangle is a right angled parallelogram. 3. A Trapczium is a quadrilateral which has two of its sides parallel. POSTULATES. 1. Let it be granted 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 are equal to one another. 7. Things which are halves of the same 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. Two straight lines which cut one another cannot both be parallel to a third straight line. PROPOSITION 1. PROBLEM. To describe an equilateral triangle on a given finite straight line. Let AB be the given straight line. It is re1. quired to describe an equilateral triangle on AB. From the centre A, at the distance AB, describe 2. the circle BCD. [Post. 3.] From the centre B, at the distance BA, describe the circle ACE. [Post. 3.] From the point C, in which the circles cut one another, draw the straight lines CA, 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. [Def. 15.] And because the point B is the centre of the.circle ACE, BC. is equal to BA. [Def. 15.] 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 equal to one another; [Ax. 1.] therefore CA is equal to CB. Therefore CA, AB, BC are equal to one another. Therefore the triangle ABC is equilateral, [Def. 24.] and it is described on the given straight line AB. Q.E.F. The letters Q.E.F. are a contraction of the words 'quod erat faciendum,' meaning which was to be done, and are usually added at the close of a problem. Ex. Describe a rhombus whose sides shall each of them be equal to one of its diagonals. PROPOSITION 2. PROBLEM. From a given point to draw a straight line equal to a given straight line. 1. 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 2. to B draw the straight line AB;[Post. 1.] and on it describe the equilateral triangle DAB, [I. 1.) and produce the straight lines DA, DB to E and F. [Post. 2.] From the centre B, at the distance BC, describe the circle CGH, meeting DF in G. [Post. 3.] From the centre D, at the distance DG, describe the circle GKL, meeting DE in L. [Post. 3.] AL shall be equal to BC. 3. Because the point B is the centre of the circle CGII, BC is equal to BG. [Def. 15.] And because the point D is the centre of the circle GKL, DL is equal to DG; [De 15.] and DA, DB parts of them are equal; (Def. 24.] therefore the remainder AL is equal to the remainder BG. [Ax. 3.] But it has been shown 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. [Ax. 1.] Therefore AL is equal to BC. Therefore from the given point A a straight line AL has been drawn equal to the given straight line BC. Q.E.F. This problem admits of eight cases in the figure. 1. The given point may be joined with either extremity of the given line. |