EXERCISES. 1. On a given straight line, to describe a segment which shall contain a given angle. SUGGESTION. On AB construct (by 187) ▲ BAD = 2 C ; draw AH perpendicular to AD at D (by 184, Ex. 1, see 151); bisect AB (by 183), and O will be the centre (see 147) and AFB the required arc (see 179). 2. Through a given point inside of a circle A other than the centre to draw a chord which is bisected at that point. SUGGESTION. Find the centre, draw the diameter through the given point, and see 147. E 0 F PROPOSITION XXVII. PROBLEM. 196. To draw a tangent to a given circle through a given point without the circumference. H B Let O be the centre of the given circle, and A the given point without the circumference. To draw through A a tangent to the circumference. It is known (from 150) that a radius and tangent drawn to its extremity are perpendicular to each other. It is also known (from 177) that a diameter subtends a right angle. Therefore the line joining the point through which the tangent is to pass and the centre of the given circle must subtend a right angle or be the diameter of an auxiliary circle. Hence draw AO, bisect it (by 183) at O', say, then describe a circle with O'O as a radius and O' a centre, and connect A with the points where this circle intersects the given circle, say B and C, then AB and AC will be the tangents required, ZABO and Z ACO being right angles. and 150. EXERCISES. 1. To describe a circle tangent to a given straight line, having its centre at a given point. SUGGESTION. See 184 2. Through a given point to describe a circle of given radius, tangent to a given straight line. с 3. Show when the above problem is impossible. E 4. To describe a circle touching two given straight lines, one of them at a given point. SUGGESTION. P с 0 C A B SUGGESTION. Erect DB (C) 1 to AB at B, draw DE parallel to AB, with P as centre, and radius equal to C, cut ED in O, then O is the centre. See 160, 150. 5. To find the centre of a given arc or circle. 6. To draw a tangent common to two circles, C and c. SUGGESTION. Draw two parallel radii CB and cb; draw Bb and continue it until it meets Cc produced in A. See 196. BOOK III. RATIO AND PROPORTION. SIMILAR FIGURES. DEFINITIONS. 197. A Proportion is an equality of ratios. That is, if the ratio of a to b is equal to the ratio of c to d, they form a proportion, which may be written 198. The four terms of the two equal ratios are called the Terms of the proportion. The first and fourth terms are called the Extremes, and the second and third the Means. Thus, in the above proportion, a and d are the extremes, and b and c the means. The first and third terms are called the Antecedents, and the second and fourth the Consequents. Thus, a and c are the antecedents, and b and d the consequents. The fourth term is called a Fourth Proportional to the other three. Thus, in the above proportion, d is a fourth proportional to a, b, and c. In the proportion a:bb: c, c is a third proportional to a and b, and b is a mean proportional between a and c. PROPOSITION I. 199. If four quantities are in proportion, the product of the extremes is equal to the product of the means. Let a: b = c: d. To prove ad = bc. By definition (197), Clearing of fractions, 200. COR. If then (by 199) ad=bc. b2 = ac. Q.E.D. That is, the mean proportional between two quantities is equal to the square root of their product. That is, α C b d Dividing both members of the equation by bd, ad bc or PROPOSITION II. THEOREM. 201. CONVERSELY, if the product of two quantities is equal to the product of two others, one pair may be made the extremes, and the other pair the means, of a proportion. Let ad = bc. = Then (by 199) a: α с d = b = c: d. a: b PROPOSITION III. THEOREM. 202. In any proportion the terms are in proportion by Alternation; that is, the first term is to the third as the second term is to the fourth. Let = c: d. Q.E.D. ad = bc. a:cb: d. Q.E.D. Q.E.D. To prove If then (by 199) PROPOSITION IV. 203. If four quantities are in proportion, they are in proportion by Inversion; that is, the second term is to the first as the fourth term is to the third. Let that is, or Divide by ac, or To prove If a: b = b: a=d: c. a: bc:d, bc = ad. bc ad ac ac d =-9 b: a=d: c. 18010 a = c: d. PROPOSITION V. THEOREM. 204. In any proportion the terms are in proportion by Composition; that is, the sum of the first two terms is to the first term as the sum of the last two terms is to the third term. Let a: a: b = c:d, then (by 199) ad = bc. Adding both members of the equation to ac, ac + ad = ac + bc, a(c + d) = c(a + b). Therefore (by 201), a+b: a::c+d: c. Similarly, a+bb::c+d: d. = c: d. Q.E.D. Q.E.D. |