Proportionality and Midpoint Theorem | IEB Grade 12 Maths

Q: What reason do I write when I use the proportionality theorem?

Write 'line $\parallel$ one side of $\triangle$' or 'prop theorem' and name the parallel lines, for example '$DE \parallel BC$'.

The proportionality theorem tells you how a line parallel to one side of a triangle divides the other two sides. It leads into similar triangles and is tested both as an examinable proof and in calculations with ratios and areas.

A line parallel to one side divides the other sides proportionally

Theorem. A line drawn parallel to one side of a triangle divides the other two sides proportionally. (Reason: line $∥$ one side of $△$ OR prop theorem; name the parallel lines)

If $D E ∥ B C$ in $△ A B C$ , with $D$ on $A B$ and $E$ on $A C$ :

$\frac{A D}{D B} = \frac{A E}{E C}$

Proof (examinable). Join $B E$ and $C D$ , and draw heights $h$ from $E$ to $A B$ and $k$ from $D$ to $A C$ .

$\frac{area △ A D E}{area △ B D E} = \frac{\frac{1}{2} A D \cdot h}{\frac{1}{2} D B \cdot h} = \frac{A D}{D B}$
$\frac{area △ A D E}{area △ C E D} = \frac{A E}{E C}$ (same method with height $k$ )
area $△ B D E$ = area $△ C E D$ (same base $D E$ , between parallel lines $D E$ and $B C$ )

So $\frac{A D}{D B} = \frac{A E}{E C}$ .

Example. $A D = 6$ , $D B = 4$ and $A E = 9$ . Then $\frac{6}{4} = \frac{9}{E C}$ , so $E C = 6$ .

The converse

Converse. If a line divides two sides of a triangle in the same proportion, it is parallel to the third side. (Reason: line divides two sides of $△$ in prop)

Example. In $△ P QR$ , $S$ is on $P Q$ and $T$ is on $P R$ with $P S = 3$ , $S Q = 6$ , $P T = 4$ and $T R = 8$ . Since $\frac{3}{6} = \frac{4}{8}$ , $S T ∥ QR$ .

Use the converse when a question asks you to prove two lines are parallel and gives you lengths or ratios.

The midpoint theorem

Theorem. The line joining the midpoints of two sides of a triangle is parallel to the third side and equal to half of it. (Reason: midpt theorem)

Converse. A line through the midpoint of one side, parallel to another side, bisects the third side. (Reason: line through midpt $∥$ to 2nd side)

Example. In $△ A B C$ , $D$ and $E$ are the midpoints of $A B$ and $A C$ . If $B C = 24$ cm, then $D E = 12$ cm and $D E ∥ B C$ .

Area ratios

Triangles with the same height have areas in the same ratio as their bases:

$\frac{area △ _{1}}{area △ _{2}} = \frac{base _{1}}{base _{2}}$

Triangles on the same base between the same parallel lines have equal areas.

Example. In $△ P QT$ , $S$ lies on $P Q$ with $P S : S Q = 2 : 3$ . Triangles $P S T$ and $P QT$ share the height from $T$ to $P Q$ , so

$\frac{area △ P S T}{area △ P QT} = \frac{P S}{P Q} = \frac{2}{5}$

Write area ratios as fractions and simplify. Always state which height is shared.

Key terms

Proportional: Two ratios are proportional when they are equal.
Midpoint: The point dividing a line segment into two equal parts.
Ratio: A comparison of two quantities, written $a : b$ or $\frac{a}{b}$ .
Same height: Triangles whose third vertices are the same distance from the line containing their bases.

Common mistakes

Writing $\frac{A D}{A B} = \frac{A E}{E C}$ , mixing a part with a whole.
Using the proportionality theorem without naming the parallel lines in the reason.
Assuming a midpoint without it being given or proved.
Comparing areas of triangles that do not share a height.

Practice questions

Reveal steps one at a time, then tick the marks you earned.

Question 1

4 marks total

routine

In $△ A B C$ , $D$ lies on $A B$ and $E$ lies on $A C$ with $D E ∥ B C$ . $A D = 6$ cm, $D B = 4$ cm and $A E = 9$ cm. Calculate $E C$ .

Worked solution

Hidden - try the question first

Solution

$\frac{A D}{D B} = \frac{A E}{E C}$ (line $∥$ one side of $△$ ; $D E ∥ B C$ )
$\frac{6}{4} = \frac{9}{E C}$
$E C = 6$ cm

Question 2

6 marks total

knowledge

Prove the theorem that states that a line drawn parallel to one side of a triangle divides the other two sides proportionally.

Worked solution

Hidden - try the question first

Solution

Given $△ A B C$ with $D E ∥ B C$ . Construction: join $B E$ and $C D$ ; draw heights $h$ and $k$
$\frac{area △ A D E}{area △ B D E} = \frac{\frac{1}{2} A D \cdot h}{\frac{1}{2} D B \cdot h} = \frac{A D}{D B}$
$\frac{area △ A D E}{area △ C E D} = \frac{A E}{E C}$
area $△ B D E$ = area $△ C E D$
(same base $D E$ ; between $∥$ lines $D E$ and $B C$ )
$\frac{A D}{D B} = \frac{A E}{E C}$

Question 3

7 marks total

complex

In $△ P QR$ , $S$ lies on $P Q$ and $T$ lies on $P R$ with $S T ∥ QR$ and $P S : S Q = 2 : 3$ . $U$ lies on $QR$ with $T U ∥ P Q$ . $QR = 20$ cm.

(a) Write down $P T : T R$ . (1)

(b) Calculate the length of $Q U$ . (3)

Worked solution

Hidden - try the question first

Solution

(a) $△ A B C$ (line $∥$ one side of $△$ ; $S T ∥ QR$ )
(b) In $△ R P Q$ : $\frac{R U}{U Q} = \frac{R T}{T P} = \frac{3}{2}$ (line $∥$ one side of $△$ ; $T U ∥ P Q$ )
$Q U = \frac{2}{5} \times QR$
$= 8$ cm
(c) $△ P S T$ and $△ P QT$ have the same height from $T$ to $P Q$
$\frac{area △ P S T}{area △ P QT} = \frac{P S}{P Q}$
$= \frac{2}{5}$

Question 4

9 marks total

problem-solving

In $△ A B C$ , $D$ is the midpoint of $A B$ and $E$ is the midpoint of $A C$ . $F$ lies on $B C$ with $B F : F C = 1 : 3$ . $A F$ cuts $D E$ at $G$ . $B C = 24$ cm.

(a) Prove that $G$ is the midpoint of $A F$ . (3)

(b) Calculate the length of $D G$ . (2)

(d) Write down $D G : GE$ . (2)

Worked solution

Hidden - try the question first

Solution

(a) $△ A B C$ (midpt theorem)
So $D G ∥ B F$ in $△$ , and $D$ is the midpoint of $A B$
$A G = GF$ (line through midpt $∥$ to 2nd side)
(b) $B F = \frac{1}{4} \times 24 = 6$ cm
$D G = \frac{1}{2} B F = 3$ cm (midpt theorem)
(c) $D E = \frac{1}{2} B C = 12$ cm (midpt theorem)
$GE = 12 - 3 = 9$ cm
(d) $D G : GE = 3 : 9$
$= 1 : 3$

FAQs

Is the proportionality theorem proof examinable?

Yes. You need the area-based proof with its construction and reasons.

What reason do I write when I use the proportionality theorem?

Write 'line $∥$ one side of $△$ ' or 'prop theorem' and name the parallel lines, for example ' $D E ∥ B C$ '.

How do area ratios work in geometry?

Triangles with the same height have areas in the ratio of their bases. Find a shared vertex and a common line for the bases.

Proportionality and the midpoint theorem

A line parallel to one side divides the other sides proportionally

The converse

The midpoint theorem

Area ratios

Key terms

Common mistakes

Practice questions

FAQs

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