where we have the same values as in the original solution, only changing + intox, as we evidently ought to do, since x is measured from A in an opposite direction. This is obvious also from the comparison of the figures 37, and 37, where all the parts of the one correspond to all the parts of the other. The figure 37, will have the same double construction that 37 has in 372. What would be the result of making p = 0? what, when p = 0, a=0? The problem we have just been discussing is admirably adapted to show the correlation of algebraical signs and geometrical figures. This problem will also enable us to divide any polygon into any required parts. 28°. The parallel sides of a trapezoid are a, b, and the altitude, h, it is required to cut off an area m2 adjacent to b by a line parallel thereto. What will be its breadth? = 29°. Given the sides of a triangle, ABC, viz., AB=10 chains, AC 8 chains 43 links, BC = 4'70; also the position of a point, P, viz., distant from AB by 1'80, and from AC by one chain. Required to draw a line through the point P, that shall divide the triangle into two equal parts. The student will solve and verify. x, then will the peri 30°. What is the greatest rectangle with a given perimeter ? We may denote the sides by a +x and a meter be 4a, and we shall find = m2 denoting the area; whence it is obvious that m2 cannot exceed a2; and therefore that = 0, when the area m2 is a maximum, or that among rectangles of the same perimeter, the square is the maximum. 31°. What is the maximum triangle that can be constructed on a given base, and of a given perimeter ? Ans. The triangle must be isosceles. 32°. Given the area and diagonal of a rectangle, to determine its sides. 33°. Let ABCD be a quadrilateral field, of which the sides AB, BC, CD, DA, are, respectively, 16, 34, 30, 29 rods in length; also, let the diagonal BD = 374 rods. It is required to divide the field into two equal parts, by a line cutting the opposite sides, AB, CD, so that the ratio of the segments of the one shall be equal to that of the corresponding segments of the other. Ans. AX = ? DY=? [Produce AB, CD, until they meet, and consult (140), (141), (132), (159).] BOOK THIRD. PLANE GEOMETRY DEPENDING ON THE CIRCLE, SECTION FIRST. The Circle. Definition 1. The circle is a plane figure described by the revolution of a straight line of invariable length about one of its extremities as a fixed point. Def. 2. The describing line is called the Radius [rod] of the circle, the fixed point the Centre, and the curve line that bounds it its Circumference. Cor. All radii, or lines drawn from the centre to the cir- (162) cumference, are equal to each other. PROPOSITION I. Angles at the centre of the same or equal circles, are to (163) each other as their subtending arcs; and the corresponding sectors have a like ratio. In the first place, let the angles be commensurable. For éxample, suppose the angle AOB to contain the angle BOC twice; then it is manifest that in applying the angle BOC twice to the angle AOB, the point C will fall on C', the middle point of the arc AB, since (162) OC′ = OC. Therefore the arc AB contains the arc BC twice; and, in the same way, the sector AOB is double the sector BOC. Fig. 38. .. ZAOB: BOC :: arc AB arc BC: sec. AOB: sec. BOC, each being as 2 to 1. KL B So in general, if a, b, be any commensurable angles, or such that when a is divided into any number, m, of equal parts, b shall also be exactly divisible into some number, n, of the same equal parts; then it will follow, by superposing one of these equal angles m times upon a and n times upon b, that the corresponding arcs A and B will be divided into m and n equal The same of the sectors K, L. arcs. Fig. 382. ab ab: A B :: K: L, being as m to n. Finally, let us take the most general case, or that of incommensurability, where a and b having no common measure, are incapable of being divided into the same equal parts. Let b be increased by the angle x, so that b+x shall be commensurable with a, then will the corresponding arc B+X be commensurable with A; and, from what has just been proved, there results the proportion Fig. 383. Cor. 1. In the same or equal circles, arcs may be taken (164) as the measures of their angles at the centre. Thus, if b be taken for the unit of angles and B for Scholium. As the right angle seems the most suitable for comparing angles, so its measure, the Quadrant, or quarter circumference, would appear to be the appropriate unit of arcs, and for this purpose the French have sometimes employed it, dividing the quadrant into a hundred equal parts, which they called degrees: but custom has established a different unit, theth part of the quad rant, which is denominated a degree, and written 1°. The degree is divided into 60 minutes, marked 1' and the minute into 60 seconds, written 1", 2", 3", .... Cor. 2. A quarter circumference is the measure of a right (165) angle, a semicircumference of two right angles, and a circumference of four right angles. How many degrees in half a quadrant? in one third of a quadrant? in ? ? ? + ? Į? Į? to? Cor. 3. In the same or equal circles, the greater arc (166) subtends the greater angle at the centre, and, conversely, the greater angle is subtended by the greater arc. Cor. 4. In the same or equal circles, equal arcs sub- (167) tend equal angles at the centre, and the converse. Cor. 5. In the same or equal circles, the greater arc sub- (168) tends the greater chord, and conversely, the greater chord is subtended by the greater arc. For, let the arc B> A, .. (166) ≤ b > a; ï. (116) chord d> c. Conversely, let chord d>c, .. < b> a; .. arc B > A. Fig. 385. Cor. 6. In the same or equal circles, equal arcs subtend (169) equal chords, and the converse. Cor. 7. The Diameter, or chord passing through the (170) centre, is the greatest straight line that can be drawn in a circle. Why? Fig. 386. Cor. 8. The diameter bisects the circle, and its circum- (171) ference. PROPOSITION II. An angle inscribed in a circle, is measured by half its (172) subtending arc. Let AOB be any diameter, and C any point of the circumference: join CA, CO, then (94), (162), (101), |