Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

THEOREM 42.

122. If four quantities, a, b, c, d, be proportionals, the first will be to the sum of the first and second as the third is to the sum of the third and fourth. For, since a, b, c, d, are proportionals, ad bc. To each side of this equation add the product ac, and we have ad+ac=bc+ac; that is, a(c+d)=c(a+b). Therefore, a a+b :: c : c+d.

123. COROLLARY.-Since out of the two equal products ad, bc, four combinations in twos may be chosen, viz.

ab, ac, bd, cd,

we may therefore have the four following equations all different, by adding each combination to each side of the equation ad = bc, viz.

No. 1....a(c+d)=c(a+b)

2....a(b+d)b(a+c)
3....d(a+b) = b(c+d)
4....d(a+c) = c(b+d)

Since each of these equations will give eight sets of proportionals; therefore the whole four will give thirty-two.

THEOREM 43.

124. If four quantities, a, b, c, d, be proportionals, as the first is to the difference of the first and second, so is the third to the difference of the third and fourth.

For, since a, b, c, d, are proportionals, ad=bc. Subtract each side of this equation from the product ac, and we have ac-ad-ac-bc; that is, a(c-d)=c(a - b), whence a: a-b::c: c-d.

125. COROLLARY.-Since out of the two equal products ad, bc, we may choose the four combinations in twos, viz. ab, ac, bd, cd; and may therefore have the four following equations, by subtracting each side of the original equation, ad=bc, from each combination.

[blocks in formation]

126. Again, by subtracting each of the products, ab, ac, bd, cd, from each side of the original equation, ad = bc, we have

No. 1....a (d-c) = c(b− a)

2....a (d-b)b(c-a)

3....d(a - b)=b(c-d)

4....d (a-c) = c(b− d)

Now, since every one of these eight different equations will give eight sets of proportionals, in each set of which the situations of the terms will be varied; therefore the whole will give sixty-four sets of proportionals, in which the terms will have different situations in every two sets.

THEOREM 44.

127. If four quantities, a, b, c, d, be proportionals, it will be, as the sum of the first and second is to their difference, so is the sum of the third and fourth to their difference.

For, dividing the equation, No. 3, pr. 123, by the equation, No. 3, pr. 126, we shall have

[blocks in formation]

And, by dividing the equation, No. 3, in 123, by the equation, No. 3, in pr. 125, we have,

[blocks in formation]

Wherefore a+ba~b::c+d:c_d*

128. Three quantities are said to be proportionals when the first is to the middle quantity as the middle quantity is to the third; thus, let a, b, c, be three proportionals; then a : b::b: c.

* The character ~ between two quantities signifies the difference, as 5—9=4.

THEOREM 45.

129. If three quantities be proportionals, the product of the two extremes is equal to the square of the mean.

Let a, b, c, be the three proportionals; then, ac = b2, for, since, multiply both sides of this equation by ab, and ab=abc, that is, b2 = ac.

THEOREM 46.

130. If the first and second terms of two proportions are alike, the third and fourth terms of both, placed in the order of their antecedents and consequents, will be proportionals.

Let a b:: cd, and a: b:: e:f; then will c: d::e: f. From the first proportion we have ad = bc, and from the second we have be = af; therefore, multiplying the corresponding sides of these equations, we have adbe=bcaf; and, throwing out the common factors on each side of this equation, there will remain de=cf; wherefore, c: d::e:f.

THEOREM 47.

131. In any number of proportionals, of which all the ratios are equal, it will be, as the antecedent of any ratio is to its consequent, so is the sum of all the antecedents of the other ratios to the sum of all the consequents.

[blocks in formation]

Whence a (d+f+h) = b(c+e+g.)

Wherefore, a b c +e+g: d+f+h, as was to be shown.

GEOMETRY, continued.

HAVING now explained so much of the principles of Algebra and Proportion as may be requisite to elucidate the subsequent propositions, we again proceed with the ELEMENTS OF GEOMETRY. The geometric definitions, &c. have been given, generally, in pages 10 to 14; but to those already explained are to be added several which follow, as it now becomes necessary that they, also, should be known.

132. EQUIVALENT FIGURES are such as have equal surfaces, without regard to their form.

133. IDENTICAL FIGURES are such as would entirely coincide, if the one be applied to the other.

134. In EQUIANGULAR FIGURES, the sides which contain the equal angles, and which adjoin equal angles, are homologous.

135. Two figures are similar, when the angles of the one are equal to the angles of the other, each to each, and the homologous sides are proportionals. 136. In two CIRCLES, similar sectors, similar arcs, or similar segments, are those which have equal angles at the centre.

[merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors]

equal to the angle DEF; or, if the arc AC be similar to the arc DF, then the angle at B will be equal to the angle at E. Also, if the segment GMH be similar to the segment KNL, the angle I will be equal to the angle R.

137. The AREA of a figure is the quantity of surface, containing a certain

THEOREM 48.

138. Parallelograms which have equal bases and equal altitudes are equal.

[blocks in formation]

the opposite sides CD and AB are equal; and in the parallelogram ABEF, the opposite sides EF and AB are equal; therefore EF is equal to CD (84)*. Again, in the parallelogram ABCD (fig. 2), the side CD is equal to AB; and, in the parallelogram ABEF, the side FE is equal to AB; therefore EF is equal to CD. Now, in fig. 1, since CD is equal to EF, add CF to both; then will DF be equal, to CE. In fig. 2, take away the common part CF, and there will remain DF equal to CE. Therefore, in each of these figures, the three straight lines AD, DF, FA, are respectively equal to the three straight lines BC, CE, EB; and, consequently, the triangle ADF is equal to the triangle BCE: therefore, from the quadrilateral ABED take away the triangle BCE, there will remain the parallelogram ABCD; and, from the same quadrilateral, take away the equal triangle ADF, and there will remain the parallelogram ABEF: therefore the parallelogram ABCD is equal to the parallelogram ABEF.

139. COROLLARY.-Every parallelogram is equal to a rectangle, of the same base and altitude.

THEOREM 49.

140. Any triangle is equal to half a parallelogram of the same base and altitude.

For the triangle ABC is equal to the triangle ACD, and the parallelogram ABCD is equal to the sum of both triangles; and, consequently, double to one of them.

F

B

E

The figures thus inserted in a parenthesis refer to a preceding or a following paragraph; as, in this instance, to 84, on page 29.

« ΠροηγούμενηΣυνέχεια »