Why This Proof Matters#
The Theorem of Pythagoras ($a^2 + b^2 = c^2$) is something you’ve used since Grade 8. But in Grade 12, you must be able to prove it using similarity. This proof appears almost every year in the final exam and is worth 6–8 marks.
1. The Setup#
Consider a right-angled triangle $\triangle ABC$ with the right angle at $\hat{C} = 90°$.
Draw a perpendicular line from $C$ to the hypotenuse $AB$, meeting $AB$ at point $D$.
This creates three triangles:
- The original large triangle: $\triangle ABC$
- A smaller triangle on the left: $\triangle ACD$
- A smaller triangle on the right: $\triangle CBD$
2. The Key Insight: All Three Triangles Are Similar#
Proving $\triangle ABC \mathbin{|||} \triangle ACD$#
| $\triangle ABC$ | $\triangle ACD$ | Reason |
|---|---|---|
| $\hat{A}$ | $\hat{A}$ | Common angle |
| $\hat{C} = 90°$ | $\hat{D} = 90°$ | Both are right angles ($\hat{C}$ in original, $\hat{D}$ by construction) |
| $\hat{B}$ | $\hat{ACD}$ | Sum of angles in a triangle |
Conclusion: $\triangle ABC \mathbin{|||} \triangle ACD$ (AAA / equiangular)
Proving $\triangle ABC \mathbin{|||} \triangle CBD$#
| $\triangle ABC$ | $\triangle CBD$ | Reason |
|---|---|---|
| $\hat{B}$ | $\hat{B}$ | Common angle |
| $\hat{C} = 90°$ | $\hat{D} = 90°$ | Both are right angles |
| $\hat{A}$ | $\hat{BCD}$ | Sum of angles in a triangle |
Conclusion: $\triangle ABC \mathbin{|||} \triangle CBD$ (AAA / equiangular)
Therefore all three triangles are similar to each other.
3. The Proof#
From $\triangle ABC \mathbin{|||} \triangle ACD$:#
Corresponding sides are proportional:
$$ \frac{AC}{AB} = \frac{AD}{AC} $$Cross-multiply:
$$ AC^2 = AB \cdot AD \quad \ldots (1) $$From $\triangle ABC \mathbin{|||} \triangle CBD$:#
Corresponding sides are proportional:
$$ \frac{BC}{AB} = \frac{BD}{BC} $$Cross-multiply:
$$ BC^2 = AB \cdot BD \quad \ldots (2) $$Add equations (1) and (2):#
$$ AC^2 + BC^2 = AB \cdot AD + AB \cdot BD $$Factor out $AB$:
$$ AC^2 + BC^2 = AB(AD + BD) $$But $AD + BD = AB$ (the two segments make up the whole hypotenuse):
$$ AC^2 + BC^2 = AB \cdot AB $$$$ \boxed{AC^2 + BC^2 = AB^2} $$This is the Theorem of Pythagoras. $\square$
4. The Converse#
The converse is also examinable:
If $AC^2 + BC^2 = AB^2$ in $\triangle ABC$, then $\hat{C} = 90°$.
Reason: Converse of the Theorem of Pythagoras.
This is used to prove that a triangle is right-angled. Calculate all three sides, check if the square of the longest side equals the sum of the squares of the other two.
5. The General “Perpendicular from Right Angle” Results#
When a perpendicular is drawn from the right-angle vertex to the hypotenuse, the following results always hold:
| Result | Formula | Reason |
|---|---|---|
| $AC^2 = AB \cdot AD$ | From $\triangle ABC \mathbin{ | |
| $BC^2 = AB \cdot BD$ | From $\triangle ABC \mathbin{ | |
| $CD^2 = AD \cdot BD$ | From $\triangle ACD \mathbin{ |
These are the three “product” results that appear in almost every Euclidean Geometry exam question.
Memory trick: Each side squared equals the hypotenuse segment it “sits on” multiplied by the full hypotenuse (for the two legs), or the two hypotenuse segments multiplied together (for the altitude).
6. Exam Strategy: The “Product” Question#
When an exam asks you to prove something like $AB^2 = AC \cdot AD$, follow this strategy:
- Identify two triangles that contain all three sides mentioned ($AB$, $AC$, $AD$).
- Prove they are similar (usually by finding two equal angles).
- Write the ratio of corresponding sides.
- Cross-multiply to get the product form.
Example#
Prove that $PA^2 = PB \cdot PC$.
- Find two triangles: $\triangle PAB$ and $\triangle PCA$ (both contain $PA$, and one has $PB$, the other $PC$).
- Show they are equiangular:
- $\hat{P}$ is common.
- Find a second pair of equal angles (often from circle geometry or parallel lines).
- Write: $\frac{PA}{PC} = \frac{PB}{PA}$ (corresponding sides of similar triangles).
- Cross-multiply: $PA^2 = PB \cdot PC$. $\square$
7. Writing Geometry Proofs: The Rules#
Every statement needs a reason#
| Statement | Reason |
|---|---|
| $\hat{A}_1 = \hat{B}_2$ | Alt $\angle$s; $PQ \parallel RS$ |
| $\triangle ABC \mathbin{ | |
| $\frac{AB}{DE} = \frac{BC}{EF}$ | Corresponding sides of similar $\triangle$s |
| $AC^2 = AB \cdot AD$ | $\perp$ from vertex of rt $\angle$ to hypotenuse |
The order of vertices matters#
If $\hat{A} = \hat{D}$, $\hat{B} = \hat{E}$, $\hat{C} = \hat{F}$, then write:
$$ \triangle ABC \mathbin{|||} \triangle DEF $$NOT $\triangle ABC \mathbin{|||} \triangle FED$ — the matching angles must be in the same position.
🚨 Common Mistakes#
- Wrong vertex order in similarity statement: If you write $\triangle ABC \mathbin{|||} \triangle FDE$ but the angles don’t match in that order, your ratios will be wrong and you lose all the marks.
- Forgetting to prove similarity first: You cannot write a ratio of sides unless you have first proven the triangles are similar. State the similarity with its reason before using the ratio.
- Mixing up the “product” results: $AC^2 = AB \cdot AD$ is NOT the same as $AC^2 = AD \cdot BD$. Draw the diagram and label carefully.
- Not stating reasons: In a proof, every single line must have a reason in brackets or next to it. Marks are split 50/50 between statement and reason.
💡 Pro Tip: The “Colour” Method#
In your rough work, use different colours (or circle/underline) for equal angles. For example:
- Circle all angles equal to $\hat{A}$ in red.
- Underline all angles equal to $\hat{B}$ in blue.
This makes it much easier to spot which triangles are similar.