Let it be required to survey the field ABCDE. We set the cross at B' in the line AE so as to make BB'A a right angle, and in the same way we find A the points C', D'; we then measure the lines AB', B'C', C'D', D'E, and the offsets B'B, C'C, DD, when surveyed. This is a good method of determining an irregular boundary, ABCDE, as that of a river. Frequently it will be better to take the offsets from A a diagonal, AC. B' Fig. 6. the field is B Fig. 62. corners cannot B Fig. 63. D Sometimes it will be preferable, especially if the be seen from each other, to measure AB then BB' at right angles to BA, then B'C at right angles to B'B, then CC' at right angles to CB', and finally A C'D at right angles to C'C. The student should now make an actual survey and draw an accurate plot of it on paper. This may be done by aid of the ruler, rightangle, and a line of equal parts, which are to be run off of any convenient magnitude by a pair of dividers. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 H +++ + PROPOSITION III. Two lines intersected by a third, making the alternate (79) angles equal to each other, are parallel. Let the lines AB, CD be intersected by EF in O and O', making Z AOO' = OO'D; then will AB, CD, be parallel. If not, let AB, CD, produced, meet on the right in I, E A B Fig. 8. then it may be shown that they will meet on the left in a point I', and consequently that two straight lines, IBAI', IDCI', may intersect in two points, which is contrary to (67). from which it follows that the figure BOO'D may be reversed and applied to the figure CO'OA; for the line OO' being reversed and applied in O'O, the line O'D will take the direction OA and therefore coincide with OA through its whole extent, and, for a like reason, OB will coincide with O'C. Hence, if the lines OB, O'D, intersected in I, the lines OA, O'C, now coinciding with them in their reversed position, would have the same point of intersection in I'. Cor. 1. Conversely, two parallel lines intersected by a (80) third, make the alternate angles equal. = For, if the lines AB, CD, being parallel, do not make [fig. 8.] the angles AOO', OO'D equal, draw A'OB' (the student will draw the line) making / A'OO' 00'D, then will (79) A'B' be parallel to CD, but, by hypothesis, AB is parallel to CD, .. through the same point O two lines AB, A'B', have been drawn parallel to the same line CD, which is contrary to (71). Cor. 2. The equality of two alternate angles determines: (81) 1o. The equality of all the other alternate angles, whether internal or external. (Where is this shown?) 2o. The equality of the external to the opposite internal on the same side of the secant line. (Why?) 3°. The sum of the internal angles, or the sum of the external, on the same side of the secant line, to be equal to two right angles. For, if the angle DO'E=AOF, adding / BOF to both sides, we have Cor. 3. If the secant line be perpendicular to one of two (82) parallel lines, it will be perpendicular to the other also. Cor. 4. Two lines parallel to the same line, are parallel to (83) each other. [How can this be shown from (82)?] Le La Cor. 5. Two angles having their sides parallel, and lying (84) in the same direction, are equal. For, producing the sides of the angles a, b, till they meet, forming the angle c, we have (81, 2°), a = c = b. 18. Fig. 82. Cor. 6. The opposite angles of a Parallelogram are equal. (85) It is scarcely necessary to remark that a parallelogram is a quadrilateral, or four-sided figure, having its opposite sides parallel. Application. Draw parallel lines upon paper with the straightedge and rightangle, and construct them in the field by aid of the cross. Fig. 83. Fig. 8. PROPOSITION IV. The sum of the external angles of a Polygon, formed by (86) producing the sides outward, is equal to four right angles. Let a, b, c, d, e, be the external angles of a polygon; through any point, either within or out of the polygon, draw lines parallel to the sides of the polygon, forming the angles a', b', c', d', e', corresponding severally to the angles a, b, c, d, e; then (84) we have La=a', b=b', c = c', d = d', e = e'; ... adding, a+b+c+d+e= a' + b' + c' + d'+e'=+. A+a+, B+b=, C+c+,... [n] but (86) a Fig. 9. α Fig. 92. ... subtracting, A+B+C+... [n] = (n − 2) (+). Cor. 1. The sum of the angles of a polygon is equal to (87) two right angles taken as many times as there are sides less two. Making successively n = 6, 5, 4, 3, we have (n-2) (+)=4(+), 3 (+), 2 (+), (+); therefore Cor. 2. The sum of the angles of a Hexagon is equal to (88) eight right angles. Cor. 3. The sum of the angles of a Pentagon is equal to (89) the sum of six right angles. Cor. 4. The sum of the angles of a Quadrilateral is equiv- (90) alent to four right angles... Cor. 5. All the angles of a quadrilateral may be right an- (91) gles; such a figure is called a Rectangle. Cor. 6. The sum of the angles of a Triangle is equal to (92) two right angles. Cor. 7. The sum of the acute angles of a Right angled (93) Triangle is equal to a right angle Cor. 8. The external angle formed by producing one of (94) the sides of a triangle, is equal to the sum of the two opposite internal angles. Let a, b, c, be the three angles of any triangle, and c adjacent to the external angle d; we have હ Fig. 93. Scholium. The angles of a triangle will be determined: 1o. When the three angles are equal; (how ?) 2o. When two angles are equal and the third known; (how?) 3°. When two angles are given. (How?) What is a hexagon? pentagon? quadrilateral? triangle? right angled triangle? EXERCISES. 10. By aid of the straightedge and rightangle, construct about the vertex of a triangle angles equal to those at the base, and do this for triangles of different forms. 2o. Through the vertex of any triangle draw a line paral- Fig. 10. lel to the base and prove (92). 3°. The same by drawing a parallel to one of the oblique sides. 4°. The same by drawing perpendiculars to the base. 5°. From any angle of a polygon, draw diagonals to all the other angles and prove (87). 6°. From any point within a polygon, draw lines to the several angles and prove the same. 7. Let fall a perpendicular from the right angle of a right angled triangle upon the hypothenuse, and prove that the three triangles thus formed are equiangular. 8°. Having drawn the sides of two angles respectively perpendicular to each other, prove that the angles are either equal or that one is what the other wants of two right angles. 9°. Prove that two triangles are mutually equiangular when the sides of the one are severally perpendicular to the Fig. 11. Fig. 12. sides of the other. Fig. 13. SECTION SECOND. Equal Polygons.-First Relations of Lines and Angles. PROPOSITION I. If, in two Polygons, excepting three parts which are (95) adjacent, viz., two angles and an included side, or two sides and an included angle, the remaining parts, taken in the same order, are severally equal, the Polygons will be equal throughout, and the excepted parts will be equal, each to each. = First. Let AB = A'B', B B', BC = B'C', < C=C', CD =C'D': then, applying the figure ABCD to A'B'C'D', let the point A fall on A' and the line AB take the direction A'B'; the point B will fall on B', since AB A'B' by hypothesis, and = A B BC will take the direction B'C', because the angle Fig. 14. B = B', by hypothesis, and the point C will fall on C'; (why?) and finally, the line CD will coincide with C'D', (why?) so that the points A and D coinciding with A' and D', the line AD will coincide with A'D', and the polygons will coincide throughout, and A will A', AD = A'D', D = D'. = = Second. Let ≤ A = A', AB = A'B', ▲ B = B', BC = B'C', ≤ C C'; then by a superposition altogether similar to that above, it may be made to appear that the polygons will coincide throughout, and that the three adjacent parts, CD, D, DA, will be severally equal to the three, C'D', ▲ D', D'A'; and it is obvious that the same reasoning may be extended to polygons of any number of sides. Q. E. D. Cor. 1. If two triangles have two sides and the included (96) angle of the one, equal to the two sides and included angle of the other, each to each, the triangles will be equal, and the remaining three parts of the one respectively equal to the remaining three parts of the other. [The student should letter the figure and explain.] Fig.142 Cor. 2. If two triangles have two angles and the included (97) side of the one, severally equal to the two angles and the included |