When two or more ratios are equal, many propositions may be proved by introducing a single symbol to denote each of the equal ratios. For example if \(\frac{a}{b} = \frac{c}{d}\), then we can assume \(\frac{a}{b} = \frac{c}{d} = k\) 

\( \Rightarrow a = bk\) and \(c = dk\), now \(\frac{{a + c}}{{b + d}}\,\,or\,\,\frac{{a - c}}{{b - d}}\) will be \(\frac{{bk + dk}}{{b + d}} \) and \(\frac{{bk - dk}}{{b - d}}= k\).

Similarly the following properties can be proved.

1. If \(\frac{a}{b} = \frac{c}{d}\), then \(ad{\rm{ }} = {\rm{ }}bc\) and \(\frac{a}{c} = \frac{b}{d}\) (Alternando)
2. If \(\frac{a}{b} = \frac{c}{d}\) \( \Rightarrow \) \(\frac{{a - b}}{{a + b}} = \frac{{c - d}}{{c + d}}\)
   (Componendo and dividendo) 
3. If  \(\frac{a}{b} = \frac{c}{d} = \frac{e}{f} = k\), then \(k\) is also equal to:
   \(\frac{{a + c + e}}{{b + d + f}} = \frac{{a + c}}{{b + d}} = \frac{{a - c - e}}{{b - d - f}} = \frac{{\sqrt {{a^2} + {b^2} + {c^2}} }}{{\sqrt {{b^2} + {d^2} + {f^2}} }}\)
   In general if
   \(\frac{a}{b} = \frac{c}{d} = \frac{e}{f} = ...... = .k\), then 
   \(k ={\left( {\frac{{p{a^n} + q{c^n} + r{e^n} + ...}}{{p{b^n} + q{d^n} + r{f^n} + ....}}} \right)^{\frac{1}{n}}}\), 
   for example if \(\frac{a}{b} = \frac{c}{d} = \frac{e}{f} = 2\), then the value of \({\left( {\frac{{9{a^7} - {9^9}{c^7} + {{11}^{11}}{e^7}}}{{9{b^7} - {9^9}{d^7} + {{11}^{11}}{f^7}}}} \right)^{\frac{1}{{14}}}}\) will be same as \({\left( {{2^7}} \right)^{\frac{1}{{14}}}}\) = \(\sqrt 2 \)
4. If \(\frac{a}{b},\,\,\frac{c}{d}\,\,\,and\,\,\frac{e}{f}\) are four unequal numbers, then \(\frac{{a + c + e}}{{b + d + f}}\) lies between minimum and maximum of all these ratios.
5. Duplicate ratio of \(\frac{a}{b}\) is \(\frac{{{a^2}}}{{{b^2}}}\) and triplicate ratio of \(\frac{a}{b}\) is \(\frac{{{a^3}}}{{{b^3}}}\)
6. Sub duplicate ratio of \(\frac{a}{b}\) is \(\frac{{\sqrt a }}{{\sqrt b }}\) and sub triplicate ratio of \(\frac{a}{b}\) is \(\frac{{\sqrt[3]{a}}}{{\sqrt[3]{b}}}\).
7. If \({a_1}x + {b_1}y + {c_1}z = {\rm{ }}0\) and \({a_2}x + {b_2}y + {c_2}z = {\rm{ }}0,\) then 
   \(\frac{x}{{{b_1}{c_2} - {b_2}{c_1}}} = \frac{y}{{{c_1}{a_2} - {c_2}{a_1}}} = \frac{z}{{{a_1}{b_2} - {a_2}{b_1}}}\)
   
8. A ratio remains unchanged if its numerator and denominator are multiplied or divided by the same number. In other words if numerator and denominator are increased in same proportion, the ratio remains constant. For example suppose the ratio is \(\frac{4}{5}\), if 4  and 5 both are increased by 10%, the ratio remains same. For example \(\frac{4}{5} = \frac{{4.4}}{{5.5}}\). 

Example: If \(a : b : c = 2 : 3 :5\) and \(c : d = 15:17\), find  \(a : b : c : d?\)

Solution: \(a : b : c  = 2 : 3 : 5\) and  \(c : d = 15 : 17\). Multiplying first ratio by 3, we get,

\(a : b : c  = 6 : 9 : 15\)

\(c : d = 15 : 17\). Combining both the ratios we get, \(a : b : c : d = 6 : 9 : 15 : 17\)

Example: A student took five papers in an examination, where the full marks were the same for each paper. His marks in these papers were in the proportion of 6 : 7 : 8 : 9 : 10. In all papers together, the candidate obtained 60% of total marks. Then, find the number of papers in which he got more than 50% marks.

Solution: Suppose the marks obtained in the five subjects are \(6x, 7x, 8x, 9x\) and \(10x\). Total marks in all the five subjects put together = \(40x\).

Suppose the maximum marks in each of the five subjects is \(m\), then

60% of \(5m = 40x\) 

or   \(m =\frac{{8x}}{{0.6}}\)=\(\frac{{40x}}{3}\)

Hence \(x = 3m/40\)

Now, marks obtained by the student in the five subjects are:

\(6\times\frac{{3m}}{{40}}\), \(7\times\frac{{3m}}{{40}}\), \(8\times \frac{{3m}}{{40}}\), \(9\times\frac{{3m}}{{40}}\), \(10\times\frac{{3m}}{{40}}\)

=\(\frac{{18m}}{{40}},\,\frac{{21m}}{{40}},\,\frac{{24m}}{{40}},\,\frac{{27m}}{{40}},\frac{{30m}}{{40}}\)

Now number of papers in which he got more than 50% is 4.

Alternate solution: Since the gap between proportion is constant hence the middle proportion can be equated with all over percentage , hence \(8x\) is equivalent to 60% or \(x\) equals 7.5% , hence \(6x ,7x, 8x, 9x, 10x\) are 45%,52.5%,60%,67.5%,75%, respectively