The Big Idea: Parallel Lines Create Equal Ratios#
Grade 12 Euclidean Geometry is fundamentally about ratios — not just shapes and angles. The central theorem is:
If a line is drawn parallel to one side of a triangle, it divides the other two sides in the same ratio.
This is the Proportionality Theorem (also called the Basic Proportionality Theorem or Thales’ Theorem in some countries). It’s the foundation for everything else in this section — similarity, the proof of Pythagoras, and most exam questions.
1. The Proportionality Theorem#
Statement#
In $\triangle ABC$, if $DE \parallel BC$ with $D$ on $AB$ and $E$ on $AC$, then:
$$\boxed{\frac{AD}{DB} = \frac{AE}{EC}}$$Reason to write in proofs: “line $\parallel$ to one side of $\triangle$” or “prop theorem; $DE \parallel BC$”
Understanding the Ratio#
The theorem says that the parallel line cuts both sides in the same proportion. If $D$ divides $AB$ in the ratio $2:3$, then $E$ divides $AC$ in the ratio $2:3$ as well.
All Valid Ratio Forms#
Because the ratios are equal, you can write them in several equivalent ways. All of these are correct:
| Form | Equation | Logic |
|---|---|---|
| Upper : Lower | $\frac{AD}{DB} = \frac{AE}{EC}$ | Standard form |
| Lower : Upper | $\frac{DB}{AD} = \frac{EC}{AE}$ | Flip both sides |
| Upper : Whole | $\frac{AD}{AB} = \frac{AE}{AC}$ | Part to whole |
| Lower : Whole | $\frac{DB}{AB} = \frac{EC}{AC}$ | Part to whole |
The golden rule: Be consistent. If you put “top over bottom” on the left, you must put “top over bottom” on the right. Never mix “top over bottom” with “bottom over top.”
Worked Example 1 — Finding a Missing Length#
In $\triangle ABC$, $DE \parallel BC$, $AD = 4$, $DB = 6$, and $AE = 3$. Find $EC$.
By the Proportionality Theorem:
$$\frac{AD}{DB} = \frac{AE}{EC}$$$$\frac{4}{6} = \frac{3}{EC}$$Cross-multiply:
$$4 \cdot EC = 6 \times 3 = 18$$$$EC = \frac{18}{4} = 4.5$$$$\boxed{EC = 4.5}$$Worked Example 2 — Using the Whole Side#
In $\triangle PQR$, $ST \parallel QR$, $PS = 5$, $PQ = 12$, and $PR = 18$. Find $PT$.
Using the “upper : whole” form:
$$\frac{PS}{PQ} = \frac{PT}{PR}$$$$\frac{5}{12} = \frac{PT}{18}$$$$PT = \frac{5 \times 18}{12} = \frac{90}{12} = 7.5$$$$\boxed{PT = 7.5}$$Worked Example 3 — Algebraic Problem#
$$\frac{AD}{DB} = \frac{AE}{EC}$$$$\frac{x}{x + 4} = \frac{3}{5}$$In $\triangle ABC$, $DE \parallel BC$, $AD = x$, $DB = x + 4$, $AE = 3$, $EC = 5$. Find $x$.
Cross-multiply:
$$5x = 3(x + 4)$$$$5x = 3x + 12$$$$2x = 12$$$$\boxed{x = 6}$$Check: $\frac{6}{10} = \frac{3}{5}\;\checkmark$
2. The Converse of the Proportionality Theorem#
The converse is equally important:
If a line divides two sides of a triangle in the same ratio, then that line is parallel to the third side.
This is how you prove lines are parallel in exam questions.
Worked Example 4 — Proving Lines are Parallel#
$$\frac{AD}{DB} = \frac{6}{9} = \frac{2}{3}$$$$\frac{AE}{EC} = \frac{4}{6} = \frac{2}{3}$$In $\triangle ABC$, $D$ is on $AB$ and $E$ is on $AC$ such that $AD = 6$, $DB = 9$, $AE = 4$, $EC = 6$. Prove that $DE \parallel BC$.
Since $\frac{AD}{DB} = \frac{AE}{EC}$, therefore $DE \parallel BC$ (converse of proportionality theorem). $\square$
3. The Midpoint Theorem#
The Midpoint Theorem is a special case of the Proportionality Theorem where the ratio is $1:1$.
Statement#
If $D$ and $E$ are the midpoints of sides $AB$ and $AC$ respectively in $\triangle ABC$, then:
- $DE \parallel BC$
- $DE = \frac{1}{2}BC$
Why It Works#
If $D$ is the midpoint of $AB$: $\frac{AD}{DB} = \frac{1}{1}$
If $E$ is the midpoint of $AC$: $\frac{AE}{EC} = \frac{1}{1}$
Since the ratios are equal, $DE \parallel BC$ by the converse of the Proportionality Theorem. The length relationship ($DE = \frac{1}{2}BC$) follows from the similarity ratio being $1:2$.
The Converse of the Midpoint Theorem#
A line drawn through the midpoint of one side of a triangle, parallel to a second side, bisects the third side.
Worked Example 5 — Midpoint Theorem#
In $\triangle ABC$, $M$ is the midpoint of $AB$ and $N$ is the midpoint of $AC$. If $BC = 14$ cm, find $MN$.
By the Midpoint Theorem:
$$MN = \frac{1}{2}BC = \frac{1}{2}(14) = \boxed{7\text{ cm}}$$Also, $MN \parallel BC$.
Worked Example 6 — Converse of Midpoint Theorem#
In $\triangle PQR$, $M$ is the midpoint of $PQ$ and $MN \parallel QR$ with $N$ on $PR$. If $PR = 10$, find $PN$.
By the converse of the Midpoint Theorem, $N$ is the midpoint of $PR$:
$$PN = \frac{1}{2}PR = \frac{1}{2}(10) = \boxed{5}$$4. Multiple Parallel Lines (Extension)#
When three or more parallel lines cut two transversals, they divide the transversals in the same ratio.
If $AB \parallel CD \parallel EF$ and two transversals cut these lines:
$$\frac{AC}{CE} = \frac{BD}{DF}$$This is a generalisation of the Proportionality Theorem and appears in harder exam questions.
Worked Example 7 — Three Parallel Lines#
$$\frac{5}{8} = \frac{3}{x}$$$$5x = 24$$$$x = 4.8\text{ cm}$$Three parallel lines cut two transversals. On the first transversal, the segments are $5$ cm and $8$ cm. On the second transversal, the first segment is $3$ cm. Find the second segment.
5. Writing Geometry Proofs — The Format#
Every statement in a geometry proof must have a reason. Here are the standard reasons for this section:
| Statement | Accepted Reason |
|---|---|
| $\frac{AD}{DB} = \frac{AE}{EC}$ | line $\parallel$ to one side of $\triangle$ / prop theorem; $DE \parallel BC$ |
| $DE \parallel BC$ | converse of prop theorem; $\frac{AD}{DB} = \frac{AE}{EC}$ |
| $DE = \frac{1}{2}BC$ | midpoint theorem |
| $N$ is the midpoint of $PR$ | converse of midpoint theorem; $M$ midpoint, $MN \parallel QR$ |
🚨 Common Mistakes#
| Mistake | Why it’s wrong | Fix |
|---|---|---|
| Forgetting to state the parallel condition | The theorem only works if lines are parallel — you must state this | Always write “because $DE \parallel BC$” in your reason |
| Mixing ratio directions | $\frac{AD}{DB} = \frac{EC}{AE}$ is wrong — top/bottom must match | Be consistent: top/bottom on both sides |
| Confusing ratio with length | Ratio $2:3$ doesn’t mean the sides are $2$ and $3$ — they could be $4$ and $6$ | Use $2k$ and $3k$ if you need actual lengths |
| Not simplifying ratios | $\frac{6}{9}$ should be written as $\frac{2}{3}$ for comparison | Always simplify before comparing |
| Forgetting the converse exists | To prove lines parallel, you need the converse, not the theorem itself | Theorem: parallel → ratio. Converse: ratio → parallel |
💡 Pro Tips for Exams#
1. The “Parallel Detector”#
Whenever you see parallel lines in a Grade 12 geometry question, there are only 3 things they could be testing:
- Angles: Alternate, corresponding, or co-interior angles
- Proportionality: Ratios of divided sides
- Similarity: Equal angles leading to proportional sides
If angles don’t help, check the ratios.
2. Label Your Diagram#
Mark all given lengths and ratios on the diagram immediately. Draw the parallel lines with arrows (→) to make them visible. This prevents you from mixing up which segments belong to which ratio.
3. The “$k$-Method” for Ratios#
If $AD:DB = 2:3$, write $AD = 2k$ and $DB = 3k$. Then $AB = 5k$. This lets you work with actual algebraic expressions instead of abstract ratios.