In mathematics, a proportion is an equation stating that two ratios are strictly equivalent. When numbers or quantities share a constant ratio, they are said to be proportional.

Depending on how many distinct terms are involved, proportions are categorised into two fundamental structures:

  • Four-Term Proportion: An equality between two separate ratios:

ab=cdora:b=c:d\dfrac{a}{b} = \dfrac{c}{d} \quad \text{or} \quad a : b = c : d

In this expression, a and d are called the extremes, while b and c are called the means. The term d is the fourth proportional to a, b, and c.

  • Continued Proportion (Three Terms): An ongoing chain of ratios where the second term repeats as the denominator of the first ratio and the numerator of the second:

ab=bcora:b=b:c\dfrac{a}{b} = \dfrac{b}{c} \quad \text{or} \quad a : b = b : c

In this relationship:

  • b is known as the mean proportional (or geometric mean) between a and c.
  • c is known as the third proportional to a and b.

Why Proportions Matter

  • Real-World Applications: In cartography and architecture, proportions maintain scale factors across technical blueprints and maps. In photographic and screen aspect ratios (such as 16:9), proportions ensure images resize without distortion. In financial analysis, the mean proportional represents the exact compound growth rate across successive multi-year periods.
  • Academic Relevance: The mean and third proportionals connect basic arithmetic ratios directly to similar triangles, Euclid's geometric theorems, and the common ratio (r) of geometric sequences.
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Core Theory

1. The Cross-Multiplication Property (Product Rule)

For any four quantities a, b, c, d in proportion:

ab=cd⟺ad=bc\dfrac{a}{b} = \dfrac{c}{d} \iff ad = bc

The Fundamental Rule of Proportion:

The product of the extremes always equals the product of the means:

Extreme1×Extreme2=Mean1×Mean2\text{Extreme}_1 \times \text{Extreme}_2 = \text{Mean}_1 \times \text{Mean}_2

From this single identity, we derive the formulas for each proportional term.

2. The Fourth Proportional

Given three known quantities a, b, and c, the fourth proportional d completes the equality:

ab=cd⟹ad=bc⟹d=bca\dfrac{a}{b} = \dfrac{c}{d} \implies ad = bc \implies d = \dfrac{bc}{a}
  • In simple words: multiply the second and third values together, then divide by the first value.

3. The Mean Proportional (Geometric Mean)

When three positive quantities a, b, and c are in continued proportion:

ab=bc\dfrac{a}{b} = \dfrac{b}{c}

Applying cross-multiplication:

b×b=a×c⟹b2=acb \times b = a \times c \implies b^2 = ac

Taking the square root of both sides gives the formula for the mean proportional b:

b=acb = \sqrt{ac}
  • The mean proportional between two numbers is the positive square root of their product.
  • In statistics and sequences, this value is identically called the geometric mean.
  • If the numbers can be signed, b=±acb = \pm\sqrt{ac}, though geometric lengths are strictly positive (b > 0).

4. The Third Proportional

In a continued proportion of three terms, the third proportional c is the final term that extends the ratio established by the first two terms a and b:

ab=bc\dfrac{a}{b} = \dfrac{b}{c}

Cross-multiplying gives:

ac=b2⟹c=b2aac = b^2 \implies c = \dfrac{b^2}{a}
  • In simple words: take the second number, square it, and divide the result by the first number.

Worked Examples

Worked Example 1: Finding the Fourth Proportional

Find the fourth proportional to 4, 9, and 16.

Step 1: Set up the proportion equation

Let the fourth proportional be d:

49=16d\dfrac{4}{9} = \dfrac{16}{d}

Step 2: Apply the cross-multiplication rule

4×d=9×164 \times d = 9 \times 16

4d=1444d = 144

Step 3: Solve for d

d=1444=36d = \dfrac{144}{4} = 36

The fourth proportional is 36.

Worked Example 2: Finding the Third Proportional

Find the third proportional to 8 and 12.

Step 1: Set up the continued proportion equation

Let the third proportional be c:

812=12c\dfrac{8}{12} = \dfrac{12}{c}

Step 2: Cross-multiply

8×c=12×128 \times c = 12 \times 12

8c=1448c = 144

Step 3: Solve for c

c=1448=18c = \dfrac{144}{8} = 18

The third proportional is 18.

Worked Example 3: Finding the Mean Proportional

Determine the mean proportional between 4 and 25.

Step 1: Set up the continued proportion equation

Let the mean proportional be b:

4b=b25\dfrac{4}{b} = \dfrac{b}{25}

Step 2: Cross-multiply

b2=4×25b^2 = 4 \times 25

b2=100b^2 = 100

Step 3: Take the positive square root

b=100=10b = \sqrt{100} = 10

The mean proportional is 10.

Practice Questions & Solutions

1

Find the fourth proportional to , , and .

Solution

Set up the proportion:

Cross-multiply:

Divide by :

2

Find the third proportional to and .

Solution

Set up the continued proportion:

Cross-multiply:

Divide by :

3

Find the mean proportional between and .

Solution

Let the mean proportional be b:

Compute the product:

Evaluate the square root:

4

Calculate the fourth proportional to , , and .

Solution

Set up the proportion:

Cross-multiply:

Divide by :

5

Determine the third proportional to and .

Solution

Set up the third proportional formula :

Evaluate the numerator:

Divide by :

6

Calculate the mean proportional between and .

Solution

Apply the mean proportional formula :

Separate the square roots:

Since and :

7

The numbers , , and are in continued proportion. Find the positive value of .

Solution

Because , , and are in continued proportion:

Cross-multiply:

latex^2 = 4 \times 36[/latex]

Divide by :

Taking the positive square root gives:

8

Find the fourth proportional to the algebraic expressions , , and .

Solution

Let the fourth proportional be d:

Cross-multiply:

Divide both sides by (for ):

9

In a right-angled triangle, the perpendicular altitude drawn to the hypotenuse divides it into two segments of length and . Calculate the length of the altitude.

Solution

By the Altitude Theorem, the altitude h is the mean proportional between the two segments:

10

Find the third proportional to and .

Solution

Let the third proportional be c:

Factorise as :

Cancel the factor :

Summarise with AI:

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Gianpiero Placidi

UK-based Chemistry graduate with a passion for education, providing clear explanations and thoughtful guidance to inspire student success.