to BE, and at the middle point G, of AB, draw GO perpendicular to AB; from their point of intersection 0, as a centre, with a radius OB, describe the arc AMB: then will the segment AMB be the segment required. AMB is measured by half of the same arc: hence, the angle AMB is equal to the angle EBD, quently, to the given angle. and conse BOOK IV. MEASUREMENT AND RELATION OF POLYGONS. DEFINITIONS. 1. SIMILAR POLYGONS, are polygons which are mutually equiangular, and which have the sides about the equal angles, taken in the same order, proportional. 2. In similar polygons, the parts which are similarly placed in each, are called homologous. The corresponding angles are homologous angles, the corresponding sides are homologous sides, the corresponding diagonals are homologous diagonals, and so on. 3. SIMILAR ARCS, SECTORS, or SEGMENTS, in different circles, are those which correspond to equal angles at the centre. Thus, if the angles A and O are A 4. The ALTITUDE OF A TRIANGLE, is the perpendicular distance from the vertex of either an gle to the opposite side, or the opposite side produced. The vertex of the angle from which the distance is measured, is called the vertex of the triangle, and the opposite side, is called the base of the triangle. 5. The ALTITUDE OF A PARALLELOGRAM, is the perpen dicular distance between two opposite sides. These sides are called bases; one the upper, and the other, the lower base. 6. The ALTITUDE OF A TRAPEZOID, is the perpendicular distance between its parallel sides. These sides are called bases; one the upper, and the other, the lower base. 7. The AREA OF A SURFACE, is its numerical value expressed in terms of some other surface taken as a unit. The unit adopted is a square described on the linear unit, as a side. PROPOSITION I. THEOREM. Parallelograms which have equal bases and equal altitudes, are equal. Let the parallelograms ABCD and EFGH have equal bases and equal altitudes: then will the parallelograms be equal. For, let them be so placed that their lower bases shall coincide; then, because they have the same altitude, their upper bases will be in the same line DG, parallel to AB HAA The triangles DAH and CBG, have the sides AD and BC equal, because they are opposite sides of the parallel ogram AC (B. I., P. XXVIII.); the sides AH and BG equal, because they are opposite sides of the parallelogram AG; the angles DAH and CBG equal, because their sides are parallel and lie in the same direction (B. L, P. XXIV.) hence, the triangles are equal (B. L, P. V.). If from the quadrilateral ABGD, we take away the triangle DAII, there will remain the parallelogram AG; if from the same quadrilateral ABGD, we take away the tritriangle CBG, there will remain the parallelogram AC: hence, the parallelogram AC is equal to the parallelogram EG (A 3); which was to be proved. PROPOSITION II. THEOREM. A triangle is equal to one-half of a parallelogram having an equal base and an equal altitude. Let the triangle ABC, and the parallelogram ABFD, have equal bases and equal altitudes: then will the triangle be equal to one-half of the parallelogram. then, because they have equal altitudes, the vertex of the triangle will lie in the upper base of the parallelogram, or in the prolongation of that base. equal to From A, draw AE parallel to BC, forming the parallelogram ABCE. This parallelogram will be the parallelogram ABFD, from Proposition, I. But the triangle ABC is equal to half of the parallelogram ABCE (B. I., P. XXVIII., C. 1): hence, it is equal to half of. the parallelogram ABFD (A. 7); which was to be proved Cor. Triangles having equal bases and equal altitudes are equal, for they are halves of equal parallelograms. PROPOSITION III. THEOREM. Rectangles having equal altitudes, are proportional to their bases. There may be two cases: the bases may be commensurable, or they may be incommensurable. 1o. Let ABCD and HEFK, be two rectangles whose altitudes AD and HK are equal, and whose bases AB and HE are commensurable: then will the areas of the rectangles be proportional to their bases. Suppose that AB is to HE, as 7 is to 4. Conceive AB to be divided into 7 equal parts, and HE into 4 equal parts, and at the points of division, let perpendiculars be drawn to AB and HE. Then will ABCD be divided into 7, and HEFK into 4 rectangles, all of which will be equal, because they have equal bases and equal altitudes (P. I.): hence, we have, ABCD : HEFK :: 7 : 4. But we have, by hypothesis, AB : HE :: 7 : 4. From these proportions, we have (B. II., P. IV.), ABCD : HEFK :: AB : HE Had any other numbers than 7 and 4 been used, the same proportion would have been found; which was to be proved. |