Mathematics • Year 9 • Unit 4 • Lesson 9
Congruence and Similarity Tests
Build fluency with the four congruence tests (SSS, SAS, ASA, RHS) and the three similarity tests (SSS, SAS, AA) — one step at a time, from a fully worked proof through guided practice to independent test-selection problems.
1. I do — fully worked example
Read every line. Each step has a short reason on the right so you can see why, not just what.
Problem. Two triangles have sides $3, 4, 5$ cm and $6, 8, 10$ cm respectively. Show that they are similar, and state which test you used.
Step 1 — Match corresponding sides (smallest to smallest, largest to largest).
$3 \leftrightarrow 6$, $\;4 \leftrightarrow 8$, $\;5 \leftrightarrow 10$.
Reason: corresponding sides must be the ones that play the same role. Match by ordering.
Step 2 — Compute each ratio.
$\dfrac{6}{3} = 2,\quad \dfrac{8}{4} = 2,\quad \dfrac{10}{5} = 2$
Reason: SSS similarity requires all three pairs of sides to be in the same ratio.
Step 3 — Conclude with the test name.
All three ratios equal $2$, so the triangles have corresponding sides in proportion.
Reason: this is exactly the SSS similarity condition.
Step 4 — State whether they are congruent too.
Sides are NOT equal (only proportional), so the triangles are similar but not congruent.
Conclusion: The triangles are similar by SSS with scale factor $k = 2$.
2. We do — fill in the missing steps
Same structure as Section 1, but with the working faded. Fill in each blank line. 5 marks
Problem. In triangles $ABC$ and $DEF$: $AB = DE = 5$ cm, $BC = EF = 7$ cm, and $\angle B = \angle E = 60°$. Show that the triangles are congruent, and name the test.
Step 1 — List the equal parts.
$AB = DE = \_\_\_$ cm (given)
$BC = EF = \_\_\_$ cm (given)
$\angle B = \angle E = \_\_\_°$ (given)
Step 2 — Check that the angle is the INCLUDED angle (between the two equal sides).
$\angle B$ is between sides $AB$ and $BC$ → it IS the included angle. ✓
Step 3 — Name the test. Two sides and the included angle equal → this is the __________________ congruence test.
Step 4 — Conclude.
$\triangle ABC \equiv \triangle DEF$ by ____________ .
Step 5 — Bonus: write the matching third side.
Since the triangles are congruent, $AC = \_\_\_\_$ (corresponding sides).
3. You do — independent practice
For each pair, name the test (SSS, SAS, ASA, RHS, or AA) and state whether you are claiming congruence or similarity. The first four are foundation (single test). The middle two are standard (decide congruent vs similar). The last two are extension (proof-style reasoning).
Foundation — name the test
3.1 Triangle $ABC$ with $AB = 6$, $BC = 8$, $CA = 10$ and triangle $DEF$ with $DE = 6$, $EF = 8$, $FD = 10$. Congruent or similar? Which test? 1 mark
3.2 Two triangles each have angles $50°, 70°, 60°$. Congruent or similar? Which test? 1 mark
3.3 Two right-angled triangles each have hypotenuse $13$ cm and one shorter side $5$ cm. Congruent or similar? Which test? 1 mark
3.4 Triangle $ABC$ has $\angle A = 40°$, $\angle B = 80°$, $AB = 6$ cm. Triangle $DEF$ has $\angle D = 40°$, $\angle E = 80°$, $DE = 6$ cm. Congruent or similar? Which test? 1 mark
Standard — decide congruent vs similar
3.5 Triangle 1 has sides $4, 6, 8$ cm; triangle 2 has sides $6, 9, 12$ cm. Are they congruent, similar, or neither? Justify with a test name. 2 marks
3.6 Triangle $ABC$ has $AB = 4$, $AC = 6$, $\angle A = 50°$. Triangle $DEF$ has $DE = 8$, $DF = 12$, $\angle D = 50°$. Are they congruent, similar, or neither? Justify with a test name. 2 marks
Extension — short proofs
3.7 In a parallelogram $ABCD$, the diagonal $AC$ is drawn. Prove that $\triangle ABC \equiv \triangle CDA$. Use the four-line structure Given / To prove / Proof / Conclusion from the lesson. 3 marks
3.8 A triangle has angles $40°, 60°, 80°$. Another triangle has angles $40°, 60°$ and a third unknown angle. Without knowing any side, can you decide whether the triangles are similar? Justify in one sentence. 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do ($\triangle ABC$ and $\triangle DEF$)
Step 1: $AB = DE = \mathbf{5}$ cm, $BC = EF = \mathbf{7}$ cm, $\angle B = \angle E = \mathbf{60}°$.
Step 2: included angle confirmed ✓.
Step 3: SAS congruence test.
Step 4: $\triangle ABC \equiv \triangle DEF$ by SAS.
Step 5: $AC = \mathbf{DF}$ (corresponding sides of congruent triangles).
3.1 — Sides 6, 8, 10 and 6, 8, 10
All three sides equal → congruent by SSS. (Both happen to be right triangles too: $6^2 + 8^2 = 10^2$.)
3.2 — Same three angles
All three angles equal → similar by AA (or AAA). NOT congruent, because we know nothing about the sides.
3.3 — Right triangles, hyp 13, one side 5
Right angle (given), hypotenuse equal ($13$), one other side equal ($5$) → congruent by RHS.
3.4 — Two angles + included side equal
$\angle A = \angle D$, $\angle B = \angle E$ and the side $AB = DE = 6$ is between them → congruent by ASA.
3.5 — Sides 4, 6, 8 and 6, 9, 12
Ratios: $6/4 = 1.5$, $9/6 = 1.5$, $12/8 = 1.5$. All three equal → similar by SSS. Sides are not equal, so not congruent.
3.6 — Two sides + included angle, but sides only proportional
$DE/AB = 8/4 = 2$ and $DF/AC = 12/6 = 2$. Same ratio AND the included angles equal ($50°$) → similar by SAS, with $k = 2$. Sides are not equal, so not congruent.
3.7 — Parallelogram $ABCD$, prove $\triangle ABC \equiv \triangle CDA$
Given: $ABCD$ is a parallelogram with diagonal $AC$.
To prove: $\triangle ABC \equiv \triangle CDA$.
Proof:
$AB = CD$ (opposite sides of parallelogram)
$BC = DA$ (opposite sides of parallelogram)
$AC = CA$ (common side)
Therefore $\triangle ABC \equiv \triangle CDA$ by SSS.
Conclusion: The two triangles are congruent, which means the diagonal of a parallelogram divides it into two congruent triangles.
3.8 — Two angles known, third triangle's angles partially known
Yes — if two angles in any triangle are known, the third is forced (angles sum to $180°$). Both triangles have angles $40°, 60°, 80°$. Similar by AA.