What is Similarity?#
Congruence (Grade 9) means two triangles are identical — same shape AND same size. Similarity (Grade 12) means two triangles are the same shape but can be different sizes — like a photocopy zoomed in or out.
When two triangles are similar:
- All corresponding angles are equal
- All corresponding sides are in the same ratio (proportion)
1. The Theorem#
$$\text{If } \hat{A} = \hat{D},\; \hat{B} = \hat{E},\; \hat{C} = \hat{F}$$$$\text{then } \frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF}$$If two triangles are equiangular (all three pairs of corresponding angles are equal), then their corresponding sides are proportional.
We write: $\triangle ABC \mathbin{|||} \triangle DEF$
The symbol $\mathbin{|||}$ means “is similar to.”
Key insight: You only need to prove two pairs of equal angles (the third follows automatically from the angle sum of a triangle = $180°$). This is the AAA (or AA) criterion.
2. Proving Similarity — The Steps#
Step 1: Identify Two Equal Angle Pairs#
Look for:
- Common angles (two triangles sharing a vertex)
- Alternate/corresponding angles from parallel lines
- Angles in the same segment (circle geometry)
- Vertically opposite angles
- Right angles (given or from $\perp$ lines)
Step 2: State the Angles with Reasons#
Write each pair of equal angles with a geometric reason:
| Statement | Reason |
|---|---|
| $\hat{A} = \hat{D}$ | common angle / alt $\angle$s; $PQ \parallel RS$ / etc. |
| $\hat{B} = \hat{E}$ | sum of $\angle$s in $\triangle$ / corr $\angle$s; $AB \parallel CD$ / etc. |
Step 3: Write the Conclusion#
$$\therefore \triangle ABC \mathbin{|||} \triangle DEF \quad (\angle\angle\angle \text{ / equiangular})$$Critical: The order of vertices in the similarity statement must match the equal angles. If $\hat{A} = \hat{D}$, $\hat{B} = \hat{E}$, $\hat{C} = \hat{F}$, then write $\triangle ABC \mathbin{|||} \triangle DEF$ — not $\triangle ABC \mathbin{|||} \triangle FED$.
Worked Example 1 — Proving Similarity#
In $\triangle ABC$, $DE \parallel BC$ with $D$ on $AB$ and $E$ on $AC$. Prove that $\triangle ADE \mathbin{|||} \triangle ABC$.
| Statement | Reason |
|---|---|
| $\hat{A} = \hat{A}$ | Common angle |
| $\hat{ADE} = \hat{B}$ | Corresponding $\angle$s; $DE \parallel BC$ |
| $\hat{AED} = \hat{C}$ | Corresponding $\angle$s; $DE \parallel BC$ |
$\therefore \triangle ADE \mathbin{|||} \triangle ABC$ (equiangular / $\angle\angle\angle$) $\square$
3. Using Similarity for Calculations#
Once similarity is proven, you can write the ratio of corresponding sides and use it to find unknown lengths.
The Ratio Rule#
From $\triangle ABC \mathbin{|||} \triangle DEF$:
$$\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF}$$How to match sides: The first two letters of each fraction come from the similarity statement:
- $AB$ (1st and 2nd letter of $\triangle ABC$) over $DE$ (1st and 2nd letter of $\triangle DEF$)
- $BC$ (2nd and 3rd) over $EF$ (2nd and 3rd)
- $AC$ (1st and 3rd) over $DF$ (1st and 3rd)
This is why the vertex order matters — it determines which sides correspond.
Worked Example 2 — Finding a Missing Side#
$$\frac{PQ}{XY} = \frac{QR}{YZ} = \frac{PR}{XZ}$$$$\frac{6}{9} = \frac{8}{YZ} = \frac{10}{XZ}$$$\triangle PQR \mathbin{|||} \triangle XYZ$ with $PQ = 6$, $QR = 8$, $PR = 10$, and $XY = 9$. Find $YZ$ and $XZ$.
The scale factor is $\frac{6}{9} = \frac{2}{3}$, so $\frac{2}{3} = \frac{8}{YZ}$:
$$YZ = \frac{8 \times 3}{2} = \boxed{12}$$$$\frac{2}{3} = \frac{10}{XZ} \quad \Rightarrow \quad XZ = \frac{10 \times 3}{2} = \boxed{15}$$Worked Example 3 — Finding a Length Using Similarity#
In $\triangle ABC$, $DE \parallel BC$, $AD = 3$, $AB = 8$, and $BC = 12$. Find $DE$.
From Example 1, we know $\triangle ADE \mathbin{|||} \triangle ABC$.
$$\frac{AD}{AB} = \frac{DE}{BC}$$$$\frac{3}{8} = \frac{DE}{12}$$$$DE = \frac{3 \times 12}{8} = \frac{36}{8} = \boxed{4.5}$$4. The “Product Proof” — The Most Important Exam Technique#
If an exam asks you to prove something like $AB^2 = AC \cdot AD$ or $PA \cdot PB = PC \cdot PD$, it is always a similarity question.
The Strategy#
- Identify two triangles that contain all the sides mentioned in the product
- Prove they are similar (find two pairs of equal angles)
- Write the ratio of corresponding sides
- Cross-multiply to get the product form
Why the Squared Term Appears#
If the same side appears in both triangles (e.g., $AB$ is a side in both $\triangle ABD$ and $\triangle ABC$), then cross-multiplying produces:
$$\frac{AB}{AC} = \frac{AD}{AB} \quad \Rightarrow \quad AB \times AB = AC \times AD \quad \Rightarrow \quad AB^2 = AC \cdot AD$$The “squared” is not special — it happens naturally when one side is shared.
Worked Example 4 — Product Proof#
In $\triangle ABC$, $\hat{C} = 90°$ and $CD \perp AB$ with $D$ on $AB$. Prove that $AC^2 = AB \cdot AD$.
Step 1 — Find two triangles containing $AC$, $AB$, and $AD$:
$\triangle ABC$ contains $AC$ and $AB$. $\triangle ACD$ contains $AC$ and $AD$.
Step 2 — Prove similarity:
| In $\triangle ABC$ | In $\triangle ACD$ | Reason |
|---|---|---|
| $\hat{A}$ | $\hat{A}$ | Common angle |
| $\hat{C} = 90°$ | $\hat{D} = 90°$ | Given; $CD \perp AB$ |
$\therefore \triangle ABC \mathbin{|||} \triangle ACD$ ($\angle\angle\angle$)
Step 3 — Write the ratio:
$$\frac{AC}{AD} = \frac{AB}{AC}$$(1st & 3rd letters of $\triangle ABC$ over 1st & 3rd of $\triangle ACD$, etc.)
Step 4 — Cross-multiply:
$$AC \times AC = AB \times AD$$$$\boxed{AC^2 = AB \cdot AD} \quad \square$$Worked Example 5 — Product Proof with Intersecting Chords#
Two chords $AB$ and $CD$ of a circle intersect at $P$. Prove that $PA \cdot PB = PC \cdot PD$.
Step 1 — Triangles: $\triangle PAC$ and $\triangle PDB$
Step 2 — Prove similarity:
| Statement | Reason |
|---|---|
| $\hat{P}_1 = \hat{P}_2$ | Vertically opposite angles |
| $\hat{A} = \hat{D}$ | Angles in the same segment (subtended by arc $BC$) |
$\therefore \triangle PAC \mathbin{|||} \triangle PDB$ ($\angle\angle\angle$)
Step 3 — Ratio:
$$\frac{PA}{PD} = \frac{PC}{PB}$$Step 4 — Cross-multiply:
$$\boxed{PA \cdot PB = PC \cdot PD} \quad \square$$5. Similarity vs Congruence — Quick Reference#
| | Similarity ($\mathbin{|||}$) | Congruence ($\equiv$) | |—|—|—| | Meaning | Same shape, different size | Same shape AND same size | | Angles | All corresponding angles equal | All corresponding angles equal | | Sides | Proportional (same ratio) | Equal (same length) | | Symbol | $\triangle ABC \mathbin{|||} \triangle DEF$ | $\triangle ABC \equiv \triangle DEF$ | | Test | AAA (or AA) | SSS, SAS, AAS, RHS | | Use | Ratio of sides → calculate lengths, prove products | Prove sides/angles equal |
🚨 Common Mistakes#
| Mistake | Why it’s wrong | Fix |
|---|---|---|
| Wrong vertex order | If $\hat{A} = \hat{D}$ but you write $\triangle ABC \mathbin{ | |
| Writing $\equiv$ instead of $\mathbin{ | ||
| Forgetting to give reasons for ratios | “$\frac{AB}{DE} = \frac{BC}{EF}$” without a reason loses marks | Write: “corr sides of sim $\triangle$s” or “corr sides; $\triangle ABC \mathbin{ |
| Only proving one angle pair | You need two pairs of equal angles (the third is automatic) | Always state two angle pairs with reasons |
| Cross-multiplying before proving similarity | You can’t use the ratio until similarity is established | Prove similarity first, then write the ratio |
💡 Pro Tips for Exams#
1. The “Product = Similarity” Rule#
Whenever you see a product ($AB^2 = ...$, $PA \cdot PB = ...$), your brain should immediately think: “I need to find two similar triangles and cross-multiply.”
2. Finding the Right Triangles#
The sides in the product tell you which triangles to use. If the product involves $PA$, $PB$, $PC$, $PD$, look for two triangles that share vertex $P$ and each contain two of these sides.
3. The Colour Method#
In your rough work, use different colours (or symbols) for equal angles:
- Circle all angles equal to $\hat{A}$ in one colour
- Underline all angles equal to $\hat{B}$ in another
This makes it easy to spot which triangles are similar.
4. Write the Similarity Statement BEFORE the Ratio#
Always write “$\triangle ABC \mathbin{|||} \triangle DEF$” first, then derive the ratio from the statement. Never try to write the ratio without the similarity statement — you’ll almost certainly get the correspondence wrong.
⏮️ Proportionality | 🏠 Back to Euclidean Geometry | ⏭️ Proof of Pythagoras