pendicular required; as it has two points each equally distant from the extremities of the line DE (53).

56. Problem. From a given point without a straight line, to draw a perpendicular to that line, when the perpendicular will come near one extremity of the line.

Let A (fig. 32) be the point from which it is required Fig. 32. to draw the perpendicular to the line BD. From B as a centre with a radius equal to the distance BA, describe an arc below BD; and from E as a centre with a radius equal to the distance EA, describe an arc cutting the first arc in A'; then from A in the direction AA', draw the line AG, which will be the perpendicular required. As B and E are two points in the straight line BG, each equally distant from A and A'; BG must be perpendicular to AA (53); that is, BGA is a right-angle.

Any other line drawn through A, must be different from AG, and therefore differently inclined to BG; but AG is perpendicular to BG; any other line must consequently be oblique to BG. We say thereforeFrom the same point without a straight line, only one perpendicular can be drawn to that line.

57. Problem. To bisect an angle, as the angle C (fig. Fig. 33. 33). Upon the sides, take CA and CB equal to each other; draw AB, and from C draw CD perpendicular to AB; CD will bisect the angle ACB (45).

58. Problem. Through a given point, A (fig. 34), to Fig. 34. draw a straight line parallel to a given straight line, BC.

From A as a centre, with a radius sufficiently great, describe the arc ED; and from E as a centre with the same radius describe the arc AF; take EG equal to AF, and through A and G draw a straight line, which will be parallel to BC: because the alternate-internal angles GAE and AEF, being measured by equal arcs, are equal (26), and consequently the two lines AG and BC, to which they refer are parallel.

59. Suppose two parallel lines, AB and CD (fig. 35), Fig. 35. to be cut by two other parallel lines, EF, GH, the space IK ML, comprehended by these parallels, is called a parallelogram. We say then-A parallelogram is a quadrilateral (or four-sided) figure of which the opposite sides are parallel. The interior angles BIK and IKM are equal to two right angles (22); and the interior angles IKM and KML, are equal to two right angles; if, therefore, IKM added to either of the two angles BIK or KML, gives the

Fig. 35. same sum, BIK must be equal to KML. In the same manner we may show that the two angles IKM and ILM are equal to each other. We perceive that, in both cases, the equal angles are opposite to each other; we therefore say--In every parallelogram, the opposite angles are equal.

Fig. 36.

60. If we draw through the vertices K and L, the straight line KL, called a diagonal of the parallelogram; it will be a straight line meeting the two parallels AB and CD; the alternate angles ILK, LKM are equal (22); and because the straight line KL meets the two parallels EF and GH, the alternate angles IKL, KLM, are equal. We have then, the two triangles IKL, KLM, of which KL is a common side, and in which the angles IKL, KLI, of the one are respectively equal to the angles KLM, LKM, of the other; the triangles are therefore equal by the third case of equal triangles; and the sides opposite the equal angles will consequently be equal, viz. IL to KM, and IK to LM. But these equal sides are parallels comprehended between parallels. From which we derive these general truths:

(1.) The diagonal of a parallelogram divides it into two equal triangles.

(2.) The opposite sides of a parallelogram are equal. (3.) Parallel lines comprehended between parallel lines are equal.

61. Suppose AB and CD (fig. 36) to be two parallel and equal lines. By joining their extremities by the straight lines AC and BD, we have the quadrilateral ABDC; and by drawing the diagonal CB, we have two triangles, ABC and CBD. Since AB and CD are parallel, and CB is a straight line meeting these parallels, the alternate angles ABC, BCD, are equal; and as AB and CD are equal, and CB a side common to the two triangles, these triangles are equal by the second case, and therefore equal in their corresponding parts; namely, the angle ACB is equal to the angle CBD; but these angles are alternate-internal with respect to the two lines AC and BD, and the secant BC; these lines are therefore parallel (23), and the figure ABDC is a parallelogram. We therefore say--When, in any quadrilateral, two opposite sides are equal and parallel, the other two sides will be equal and parallel; and the figure will be a parallelogram.

62. As two adjacent angles of a parallelogram are

equal to two right angles, the sum of all the angles is equal to four right angles. And if one of the angles of a parallelogram is a right angle, they will all be right angles; and the figure will be what is called a rectangle.

If two adjacent sides of a parallelogram are equal; as the sides opposite are equal to them, the sides will all be equal. In this case, if the angles are right-angles, the figure is called a square. If the sides of a parallelogram are equal, and the angles oblique, the figure is called a rhombus. If the sides are unequal and the angles oblique the figure is called a rhomboid.

63. If in the rhomboidal parallelogram ABDC (fig. 37) Fig. 37. we draw the diagonals AD and CB, and compare the triangles AEB, CED, we shall have the two angles BAE, EDC, alternate-internal with reference to the parallels AB and CD; they are therefore equal; the two angles AEB and CED, are vertical angles (16), and therefore equal; consequently the third angle ABE, of the one, is equal to the third angle ECD, of the other (35); and the side AB is equal to the side CD, being opposite sides of the parallelogram. We have, therefore, the side AB, and the adjacent angles A and B, in the one triangle, equal respectively to the side CD and the adjacent angles D and C of the other; the triangles are consequently equal by the third case; and their corresponding parts are equal; namely, AE is equal to ED, and CE is equal to EB; that is, these diagonals bisect each other at E. It is evident that this fact results from the equality and parallelism of the opposite sides; whence we have the general truth-In every parallelogram, the two diagonals bisect each other.

64. If the parallelogram ABDC were a square or a rhombus, ABD would be an isosceles triangle (34), and BE would be perpendicular to AD (45). Hence-In a square or in a rhombus, the two diagonals bisect each other at right angles.

65. A trapezoid is a quadrilateral which has two of its sides parallel, and the other two oblique, to each other. If in a quadrilateral no two of its sides are parallel, the figure is called a trapezium.

Remark. Rectilinear figures of five sides are called pentagons; of six, hexagons; of seven, heptagons; of eight, octagons, &c. But all rectilinear figures are comprehended under the general name of polygons. The

Fig. 33.

Fig. 39.

sum of all the sides of a polygon is called its perimeter. 66. Problems. (1.) Having two contiguous sides and the contained angle, in a parallelogram, to construct the parallelogram.

Let A and B (fig. 38) be the given sides, and C the given angle. Draw DE equal to A; and at D, make the angle EDF, equal to the given angle C; and make DF equal to B; then through F draw FG parallel to DE, and through E draw EG parallel to DF, and the parallelogram is completed.

(2.) Construct a square upon an assumed line, as one

of its sides.

(3.) Construct a rhombus upon an assumed line, and having an assumed angle.

These examples are sufficient to enable us to construct any rectilinear figure, when we have the sides given, and the angles which they make with each other. In this way the plan of a town or an estate is constructed.

67. We have seen that the sum of all the angles in a parallelogram is equal to four right angles (62). It is desirable to know whether there is any general law respecting the angles of polygons.

Let us take the polygon ABCDE (fig. 39); produce each of its sides outward in succession; and from some point in the interior of the polygon, draw lines parallel to each of the sides. It is evident that each of the angles, a, b, c, &c. must be equal to the corresponding exterior angles of the polygon (24); but the sum of all the angles made by lines diverging from the same point in a plane, is equal to four right angles (15). It is manifest that how many sides soever the figure may have, from the same point lines may be drawn respectively parallel to each of the sides. We say therefore--The sum of all the exterior angles of every polygon, made by producing the sides outward in succession, is equal to four right angles.

68. It will be perceived, that at each vertex of the polygon, the sum of the interior and exterior angles is equal to two right angles (13); consequently the sum of all the interior and exterior angles is equal to as many times two right angles as the polygon has sides. But the sum of all the exterior angles is equal to four right-angles (67); therefore-The sum of all the interior angles of any convex polygon, is equal to as many times two rightangles as the figure has sides, wanting four right-angles.

Remark. These two propositions are limited to convex polygons; or such as have all their angles salient, that is, having the vertices of the angles outward. In polygons which have re-entering angles, that is, angles whose vertices are towards the interior of the polygon, as the angle C (fig. 40), these propositions will not be Fig. 40.

true.

69. It is evident that any polygon, ABCDEF (fig. 41), may be divided into triangles, by drawing from Fig. 41. one of the vertices, as A, diagonals to each of the other vertices not adjacent. And it is manifest that if, in two equal polygons (that is, in two polygons which would coincide if placed the one upon the other, which is called super-position), diagonals be drawn from corresponding vertices in the two polygons, to the opposite vertices, the triangles in the one polygon, would be equal respectively to the triangles in the other; and the triangles being equal, and similarly disposed, the polygons which they compose must be equal. That is-Two equal polygons may be divided into the same number of triangles, respectively equal. And--If two polygons are composed of the same number of triangles, equal and similarly disposed, the polygons are equal.

70. A more important method of determining a polygon by triangles, is that in which the triangles have for a common base, one of the sides of the polygon, as AB (fig. 42), and their summits respectively in the opposite Fig. 42. vertices of the polygon. It is manifest that the triangles ABC, ABD, ABE, ABF, having AB for a common base, determine the points C, D, E, F; all the vertices of the polygon are therefore determined, and consequently the polygon itself.

71. In this way surveyors sometime determine the most important points in a field or an estate, whether on its border or in the interior. A straight line is measured, from each extremity of which all the points to be determined can be seen. This line is called the base line, because it is made the base of a series of triangles whose summits are to determine the exact position of the other points. No point can in this way be determined, which is in the base line produced. This method of determining the points in a survey, is called triangulation. 72. We have discussed the general properties of individual triangles, and their relative parts. We have also