radii OA and OC. Because ABCD, therefore + AB AECD=CF; OA=OC; therefore the rightangled triangles OEA and OFC are equal, (135 ;) therefore OE=OF. 2. Let OE OF. In the right-angled triangles OEA and OFC we have OAOC, and OEOF; therefore triangle OEA OFC; therefore AECF. But AE=} AB, and CF = CD; therefore AB=CD. : = 236. Of two unequal chords the greater is at the least distance from the centre. 2 2 Dem. (Fig. 130.) Let HI>CD; therefore GI> CF. Now GI2+GO2=OI2, and CF2+OF2=OC2, (220.) Now Ol2=OC; therefore GI + GO2= OI 2 2 = CF2 +OF2. But GI2>CF; therefore OG <OF2; there fore OGOF. 237. Cor. The diameter is the greatest chord in a circle, because it is at no distance from the centre. 238. To describe a circle whose circumference shall pass through two given points, as A and B, (fig. 129.) Sol. Draw the straight line AB; this line will be a chord of the required circle. Bisect AB, and at the middle point G erect a perpendicular. A circle described from any point of this perpendicular with a radius equal to the distance between such point and A or B, will pass through the points A and B. 239. To describe a circle whose circumference shall pass through three given points not in the same straight line, as A, B, and C, (fig. 131.) Sol. Draw the straight lines AB and BC, which will 13 be chords of the required circle. At the middle of each of these lines erect a perpendicular. Both of these perpendiculars will pass through the centre of the required circle, (169,) which must therefore be the point O at which they meet. A circle described from this point with a radius equal to OA or OB, will be the one required. 240. We have here solved another problem, namely, to describe a circle about a given triangle. 241. To describe a circle about a regular polygon ABCDE, (fig. 97.) Sol. The sides of the given polygon will be chords of the required circle. Find the centre of this circle by erecting perpendiculars upon the middle points of these chords. This centre being in each perpendicular is at the same distance from the extremities of each chord. This distance is therefore the radius of the required circle. 242. To describe a circle whose circumference shall touch both the sides of a given angle. Sol. Let A be the angle, (fig. 132.) In the two sides take AC=AD, and at the points D and C erect perpendiculars which will intersect each other at O. From O as a centre, with a radius OC, describe a circle, which will be the circle required. Dem. Draw OA. =AO; angle ACO = = By construction AC AD; AO ADO being right angles; therefore triangle ACO ADO, therefore OC=OD. Therefore a circle described from O with a radius OC will pass through D and C; and since the radii OC and OD = are perpendicular to the sides of the angle, these sides will be tangents to the circle, (170.) = = 243. Remark. Since triangle ADO ACO, therefore angle DAO CAO; that is, an angle is bisected by a line drawn from its vertex to the point of intersection of perpendiculars erected upon its sides at equal distances from its vertex. 244. To inscribe a circle in a triangle. Sol. (Fig. 133.) Bisect the angles A and B by straight lines, which will intersect at O. From O let fall upon the three sides of the triangle, the perpendiculars OD, OF, and OE. A circle described from O as a centre, with a radius OD, OF, or OE will be the circle required. Dem. Angle ADO AFO; angle DAO-FAO; AOAO; therefore triangle DAO=FAO; therefore OD OF. Again, angle ODB=OEB; angle OBDOBE; therefore DOB=BOE, (123;) OB=OB; therefore triangle OBDOBE; consequently OD OE. Thus OD=OE OF, and a circle described from O, with a radius OD, will pass through the points D, E, and F. The radii are perpendicular to the sides of the triangle at these points, therefore the sides are tangents to the circle, (170.) 245. To describe a circle whose circumference shall touch the circumference of another circle at a given point A, and shall also pass through a given point C. (fig. 134.) Sol. Draw OA, and produce it indefinitely. Draw AC and bisect it at B. At B erect a perpendicular = which will intersect OA produced at D. The point D is the centre of the required circle, and DA is a radius. Dem. By construction BA BC; BD=BD; angle ABD CBD; therefore triangle ABD = CBD ; therefore ADCD; therefore a circle described from D as a centre, with radius DA, will pass through points A and C, and is therefore the circle required. VII. OF PROPORTIONS. 1. OF GEOMETRICAL PROPORTIONS IN GENEral. 246. The term ratio is used to denote the comparative magnitude of two quantities; for example, if one line is 2 inches long, and another line is 12 inches long, we say the lines are to one another in the ratio of 2 to 12, or of 1 to 6; that is, one is six times as great as the other. Two equal ratios form a proportion. If any quantity is contained in a second quantity, as often as a third is contained in a fourth, these four quantities form a geometrical proportion. For example, (fig. 142;) in the triangles ABC and DEF, let DE 2 AB, and DF 2 AC, and EF2 BC, we should have these propor = tions; DE: AB EF : BC EF: BC= DF: AC DE: AB=DF: AC which we should read thus; = DE is to AB as EF is to BC, &c., meaning that the quotient of DE divided by AB is equal to the quotient of EF divided by BC. The first term of each ratio is called the antecedent, the second term is the consequent. The first and fourth terms of a propor tion are called the extremes; the second and third terms are the means. If the same quantity is taken twice as a mean it is called a mean proportional between the extremes. A proportion is the equality of ratios; hence two or more ratios are required to form a proportion. Proportion does not require the equality of the quantities, but the equality of the ratios; hence two lines may form a proportion with two surfaces; for one line may be contained as often in the other line, as one surface in the other surface. 247. As a ratio expresses the number of times that one quantity is contained in another, it may be represented in the form of a fraction. Let us substitute numbers for lines, . Now let us reduce these fractions to a common denominator and we shall have 4 X 12 8 × 6 Now omit the common denominator, 6 × 12 12 × 6° = and we have 4 × 128 × 6; that is, the product of the extremes is equal to the product of the means. This proposition is true of all proportions, and may be used as a test to ascertain if a proportion exists. 248. If all the terms are of the same kind, for example, all lines, or all numbers, we may transpose the means, and yet keep the proportion; thus, 4:68:12 gives the new proportion 4:8=6:12, by transposition, as it is called. Or, we may get a new proportion by inversion, that is, by changing the places of the terms of each ratio; thus, 6:4 12:8. There are other changes which may be made without destroying the proportions, and they will be mentioned as occasion requires their use. = |