...о с— a V. Alternation ; that is, a will be to с as b is to d. 282. In a Series of Equal Batios, the sum of the antecedents is to the sum of the consequents as any antecedent is to its consequent. For if aff-2, rui, 11 — — — i bdfh r may be put for each of these ratios. Then - = r - = r -...

...Let a:b = c:d. By division, ., a—bc—d or, a-\-b:a — b= c-\-d: c — d, PROPOSITION IX. QED 303. In a series of equal ratios, the sum of the antecedents...antecedent is to its consequent. Let a: b = c: d — e :/= g: h. To prove a+c+e+g : b+d+f+ h = a: b. Denote each ratio by r. '-iH-1 Whence, a = br, c = dr,...

...application of their methods. For instance, under the theory of proportion, it is sometimes stated that : " In a series of equal ratios the sum of the antecedents...consequents as any antecedent is to its consequent." This is a true proposition applied to numbers, but is not true of geometrical magnitudes unless these...

...163), íSr;-7?H-í»HTherefore, a + c + e + g :l+d+f + h::a:b. Hence, XI. In a continued proportion the sum of the antecedents is to the sum of the consequents as any antecedent is to its consequent. a2 + Ь* а Ъ + b с EXAMPLE 1. .If ~ï~v~î~ = ~j,z 4. г > Prove that о as a mean proportional...

...ос Multiplying by -, — = — -. с ос cd ab or - = -у с d .'. a : с = b : d. 317. In a aeries of equal ratios, the sum of the antecedents is to...consequents as any antecedent is to its consequent. •c, -ta с ea For'lf ¿=5=7=f' r may be put for each of these ratios. Then |=г,.=^=г,j,.r. .-....

...ma : mb = гtc : nd ; (ii.) ma : nb= me: nd. The proof is left as an exercise for the student. 224. In a series of equal ratios, the sum of the antecedents is to the киm of the consequents as any one antecedent is to its consequent. For assume a: b = c: d= e:f= ...,...

...mb = nc : nd ; (ii.) ma : nb = me : nd. The proof is left as an exercise for tl1e student. 224. L1 a series of equal ratios, the sum of the antecedents is to the кит of the consequents as any one antecedent is to its consequent. For assume a:b = c: d = e:f=...

...= d = = д • Therefore, a + c + e + rj : b + d+f+h :: a : b. Hence, XI. In a continued proportion the sum of the antecedents is to the sum of the consequents as any antecedent is to its consequent. a2 + b3 ab + Ь с EXAMPLE 1. .If — r-^_- j- = -rj-x~-j-, prove that b is a mean proportional between...

...conversely, the greater angle is opposite the greater side. 2.- Show that in any continued proportion, the sum of the antecedents is to the sum of the consequents as any antecedent to its corresponding consequent. 3. Prove that, in equal circles, incommensurable angles at the centre...

...By division, a±bc + d a — o c—d or, a + i : « — b = c-\-d: c — d. QED PROPOSITION IX. 303. In a series of equal ratios, the sum of the antecedents...consequents as any antecedent is to its consequent. Let a: b = c:d = e:f=g:ti. To prove a + e + e + <7 : b-\-d-\-f-\- h — a : b. Denote each ratio by r. Then...