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and their included angle respectively equal. If A B F32 (fig. 32)=D E, AC=DF, and the angle B A C=E D
F, then we say the triangles A B C and D E F are equal in all their parts. This may be proved by superposition. Place D E upon A B, and by hypothesis they must coincide. Also, since the angle A=D, D F will take the same direction as A C, and since they are equal in length, the point F will fall on C, as E did on B. Then E F and B C, having two points common, cannot differ. The two triangles, therefore, coincide throughout. Accordingly we
. say-two sides and their included angle determine the triangle- ; for while these three parts do not vary, the other three, namely, the remaining side and the two remaining angles, cannot vary.
54. —Having two sides of a triangle and their included
angle given, to construct the triangle, Draw D F=to one F32 of the given sides (fig. 32). Make the angle D=to the
given angle. This determines the direction of D E, and its length, which is given, determines the point E. Thus we have two points E and F, which determine the length and position of the side EF. The triangle is therefore constructed. For, by the preceding proposition, there can be no triangle different from D E Ė, which has the same three parts given.
- 55. — Two triangles are equal, when they have a side and
two adjacent angles respectively equal — If the side A F32 B=D E (fig. 32), the angle A=D, and the angle B=E,
then we say the two triangles are equal in all their parts. The proof is by superposition as before. A B will coincide with D E by hypothesis. B C will take the same direction as E F, because the angle B=E, and therefore the point C must fall somewhere in E F. Again AC will take the same direction as D F, because the angle A=D, and therefore the point C must fall somewhere in DF. Now since C is to be in both the lines D F and E F at the same time, it can only be at their intersection F. Thus the two triangles coincide throughout. Hence we say-a side and its two adjacent angles determine the triangle
56. ---Having a side and its two adjacent angles given, to F32 construct the triangle— Draw A B (fig. 32) equal to the
given side. At X make an angle equal to one of the given angles. This will determine the direction of A C. At
B make an angle equal to the other given angle. This will determine the direction of B C. The meeting of A C and B C determines the triangle, since, by the preceding proposition, no triangle having the same parts given can differ from the one constructed.
57. -Two triangles are equal, when their three sides are respectively equal— If A B=D E (fig. 33), A C=D F, 33 and B C=E F, then we say the two triangles are equal. Place A C upon D F. Then it is only necessary to prove that the point B will fall upon E. Take D as a centre and with a radius D E make an arc at E. B must fall somewhere in this arc, because A B=D E. Again take F as a centre and with a radius F E make another arc cutting the
first. The point B must fall somewhere in this arc also, • because B C=E F. Then it can only be at their inter
section E, and the triangles must coincide throughout. Hence we say—the three sides determine the triangle
58. -Having three sides of a triangle given, to construct the triangle— Draw. D F (fig. 33) equal to one of the F33 given sides.
Take D as a centre and with a radius equal to another of the given sides, make an arc E. Again take F as a centre and with a radius equal to the remaining side, make another arc cutting the first. Then draw D E and F E, and the triangle is constructed; since, by the preceding proposition, no triangle having the same things given, can differ from the one in question.
59. -Two right triangles are equal, when they have the hypothenuse and another side equal. If B C=Ě F (fig 34), F34
“ =Ě and A C=D F, then we say the two right triangles are equal. Place A C upon its equal D F. Then A B will take the direction of D E, because A and D are right angles, and the point B will fall somewhere in D E. We wish to prove
that it will fall on E. Take F as a centre and with a radius F E make an arc cutting D E in E. B must fall somewbere in this arc,
because BC=EF. Now since B must be at the same time in the line D E and in the are, it can only be at their intersection E, and the two triangles coincide throughout. Hence we say-the hypothenuse and a side determine a right triangle
60. -Having the hypothenuse and another side given, to construct a right triangle“. Draw D F equal to the given side, and erect a perpendicular at D. Then take F as a
centre and with a radius equal to the given hypothenuse, make an are E cuing the perpendicular. Draw F E and the triangle is coustiucted, slice, by the preceding proposition, no right triangle having the same things given can diifer from the one in question.
61. It will be seen from the eight preceding articles, that--in vrder to construct a triangle or corwince ourselves of its equality to another triangle, we must always know three of its six parts, of which one at leasi must be a side. Three angles alone are not sufficient to determine a triangle. Why ?-Because any number of different triangles may be
constructed, all having their three angles respectively F35 equal. Thus the triangles A B C and DEF, (fig. 35) having
their sides parallel, are equiangular with respect to each other (39). That is A=D, B=E, and C=F. Yet the triangles are not equal; and it is evident that the number might be increased to any extent, and the same would be true.
62. We have already shown (8) that the ratio of two straight lines is expressed in the same manner and has the same meaning, as that of two abstract numbers.
We now remark that—two equal ratios in lines as well as in numbers, make a proportion- Hence the phrase geometrical proportion. To explain the nature and laws of proportion belongs to arithmetic and algebra. We shall not therefore enter into a particular analysis of them here. But for the sake of those who may not have studied proportion elsewhere, we shall briefly state the principles to be made use of hereafter. We shall illustrate the application of each by one example in numbers, since we have already shown that the value of straight lines may be represented by numbers (7). The ratio between two numbers is expressed in the form of a fraction. Thus the ratio of 6 to 9 is... A proportion expresses the equality of two ratios. Thus the equation
is a proportion. But the usual form of writing it is 6:9::10:15. This is read, 6 is to 9 as 10 is to 15, and the meaning is, that 6 is the same part of 9 that 10 is of 15. The first term in each ratio is called an antecedent, and the second a consequent.
Thus 6 and 10 are antecedents, and 9 and 15 consequents. The first and fourth term of a proportion are called extremes, and the second and third means. Thus 6 and 15 are extremes, and 9 and 10 means. If the same number is taken twice as a mean, it is called a mean proportional. Thus in the proportion 2:4::4:8, we say 4 is a mean proportional between 2 and 8. If more than two equal ratios are written after one another, they form a continued proportion. Thus 6:9::10:15::8:12 is a continued proportion.
63. -- In every proportion, the product of the means is equal to the product of the extremes For if two equal fractions be reduced to a common denomination, their numerators must be equal. Thus from the proportion 6;9::10:15 we have 9x10=6X15. This property being universal,
= furnishes a convenient test by which to ascertain the truth of a proportion, for any four numbers will be in proportion when they satisfy this condition.
64. --If two proportions have one ratio common, the other two ratios make a proportion. For ratios are nothing more than fractions, and two fractions, each of which is equal to a third, are equal to each other. Therefore these two make a proportion. Thus if we have 6:9::10:15 and 6:9::16:24, then we say 10:15:: 16:24. Apply the test and this last proportion will be found true.
65. - In every proportion the means, or the extremes, or both, may change places. For this does not affect the equality of the product of the means to that of the extremes. Thus the proportion 6 : 9::10: 15 may be written in the three following forms; 6:10::9:15; 15:9::10:6; 15:10::9:6. Apply the test and all these will be found true. 66. -In every proportion, either ratio or both ratios
be multiplied or divided by the same number, without destroying the proportion — For ratios are fractions, and to multiply or divide the numerator and denominator of a fraction by the same number does not alter its value. Thus from the proportion 6:9::10:15 we have : : 10 X2 : 15 X2. Apply the test and this last proportion will be found true.
67. –Erery proportion may be multiplied by itself or by another, term by term, and the squares or products will form a new proportion. For if two equal fractions be multiplied by two equal fractions, the products must evidently be
equal fractions, that is equal ratios, and therefore a proportion. Thus if we have 2:4::6:12 then we say 2x2: 4X4::6X6 : 12x12. Again if we bave the two proportions 2:4::6:12 and 6:9::10:15 then, multiplying term by term, we say 2X6 : 4X9:36X10:12X15. Apply the test and both will be found true.
68. every proportion, the sum of the two first terms is to that of the two last, and the difference of the two first is to that of the two last, as the first is to the third, or as the second is to the fourth. Thus from the proportion 2:4::6: 12 we have 2+4:6+12::2:6 and 4–2:12–6::4:12. Apply the test and both will be found true.
69. — In every continued proportion, the sum of any number of antecedents is to the sum of the same number of consequents, as one antecedent is to its consequent—. Thus if we have the continued proportion 2:4::6:12::8:16 then we say 2+6+8:4+12+16::2:4. Apply the test and this will be found true. The same might be proved of any number of equal ratios.
Of Proportional Lines.
70. We are now prepared to demonstrate the following proposition, upon which more depends, than upon any other in geometry. - If a line be drawn through two sides of a triangle parallel to the third side, it dirides those two sides proportionally. As the demoristration of this proposition is long, we shall divide it, for the sake of clearness, into
1. If one of the two sides be divided into any number of equal parts, and if through the points of division lines be drawn to meet the other, parallel to the third side, which we shall call the base, these parallel lines will divide the other side into the same number of equal parts. 2. If a point be taken in one side of a triangle such that the entire side shall have to the part cut off a ratio wbich can be expressed hy whole numbers, and if through this point a line be drawn parallel to the base to meet the other side, this other side will have to the part cut off the same ratio, and the sides will be divided proportionally. 3. If a point be tåken any where in the side of a triangle, and if a line be drawn througb it parallel to the base, the two sides will be divided proportionally. These three propositions we shall demonstrate in their order. First we say that if A P