Mathematics • Year 9 • Unit 4 • Lesson 9
Congruence and Similarity — Mixed Challenge
Pull together every test from Lesson 9: SSS, SAS, ASA, RHS congruence and SSS, SAS, AA similarity. You'll choose the right test for each problem, spot a flawed proof, and tackle an open-ended challenge involving the "ambiguous case" of SSA.
1. Mixed problems — choose the right test
For each pair, identify congruence vs similarity (or neither) and name the specific test (SSS, SAS, ASA, RHS, AA). Decide before you start writing. Show your working. 3 marks each
1.1 $\triangle PQR$ has $PQ = 7$, $QR = 9$, $\angle Q = 35°$. $\triangle XYZ$ has $XY = 7$, $YZ = 9$, $\angle Y = 35°$. Congruent or similar? Which test?
1.2 Two right-angled triangles each have a $90°$ angle. The first has hypotenuse $26$ and one short side $24$. The second has hypotenuse $13$ and one short side $12$. Congruent, similar, or neither?
1.3 $\triangle ABC$ has angles $30°, 60°, 90°$, side $AB = 6$. $\triangle DEF$ has angles $30°, 60°, 90°$, hypotenuse $12$. Are the angle-side combinations enough to claim congruence or only similarity?
1.4 In $\triangle ABC$, $D$ is the midpoint of $AB$ and $E$ is the midpoint of $AC$. Prove that $\triangle ADE \sim \triangle ABC$, giving the test used.
1.5 Two isosceles triangles each have a vertex angle of $40°$. Are they similar? Are they necessarily congruent? Justify.
1.6 In $\triangle ABC$, the segment $DE$ joins points on $AB$ and $AC$ so that $DE \parallel BC$. Prove $\triangle ADE \sim \triangle ABC$. State the test used.
2. Find the mistake
Another student has tried to prove that two triangles are congruent. Their working is shown below. Exactly one line contains a logical mistake (it is not an arithmetic error — it is about choosing the right test). Spot it, explain why it's wrong, then re-do the conclusion correctly. 3 marks
Student's working — claim $\triangle ABC \equiv \triangle DEF$:
Line 1: $\angle A = \angle D = 60°$ (given)
Line 2: $\angle B = \angle E = 80°$ (given)
Line 3: $\angle C = \angle F = 40°$ (angles sum to $180°$)
Line 4: Therefore $\triangle ABC \equiv \triangle DEF$ by AAA.
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write a corrected Line 4 that says exactly what AAA does prove. Then state what extra piece of information would upgrade the conclusion to congruence.
Stuck? Revisit lesson § "Common Misconceptions" — the first misconception listed is exactly this trap.3. Open-ended challenge — the ambiguous case
This question explores why "SSA" (two sides + a non-included angle) is not a valid congruence test. 4 marks
3.1 A friend insists "if two sides and any angle are equal, the triangles must be congruent." You want to disprove this with a counter-example involving SSA.
Construct a counter-example: find two different triangles that both have sides $AB = 8$ and $BC = 5$, and angle $\angle A = 30°$ (an angle that is NOT between $AB$ and $BC$ — it's at $A$).
(i) Sketch both triangles.
(ii) Briefly explain why two different triangles can satisfy these conditions (think about where point $C$ could land).
(iii) State which valid test SSA is sometimes mistaken for, and what makes the valid versions different (e.g. why SAS works but SSA doesn't, and why RHS works for right-angled triangles even though it has the "non-included" flavour).
How did this worksheet feel?
What I'll revisit before next class:
1.1 — $PQ = XY$, $QR = YZ$, $\angle Q = \angle Y$
Two sides equal, and the angle is the included angle at $Q$ (between $PQ$ and $QR$) → congruent by SAS.
1.2 — Right triangles, hyp 26/24 vs 13/12
Sides 26, 24 vs 13, 12. Ratios $26/13 = 2$, $24/12 = 2$. Both right-angled. → similar by SSS (or "RHS similarity") with $k = 2$. Not congruent.
1.3 — Same angles, different sides
$AB = 6$ in the first; hypotenuse $= 12$ in the second. With angles $30°, 60°, 90°$, side $AB$ is opposite $60°$ if it's not the hypotenuse — but the second triangle's hypotenuse is $12$. In any case the side lengths don't match, so → similar by AA, not congruent. (Need a matching side to claim congruence.)
1.4 — Midpoints $D$, $E$ → $\triangle ADE \sim \triangle ABC$
$AD/AB = 1/2$ and $AE/AC = 1/2$ (midpoints). $\angle A$ is common to both triangles. Two sides in proportion with included angle equal → similar by SAS.
1.5 — Two isosceles triangles, vertex angle $40°$
Base angles each $= (180° - 40°)/2 = 70°$, so both triangles have angles $40°, 70°, 70°$. → Similar by AA. NOT necessarily congruent — they could be different sizes.
1.6 — $DE \parallel BC$ in $\triangle ABC$
$\angle ADE = \angle ABC$ (corresponding angles, $DE \parallel BC$) and $\angle A$ is common. Two pairs of equal angles → similar by AA.
2 — Find the mistake
(a) The mistake is on Line 4.
(b) AAA only proves similarity, not congruence. Triangles with all three angles equal can be different sizes (e.g. two equilateral triangles of side $5$ cm and $8$ cm). The lesson's "Common Misconceptions" lists this as the first pitfall.
(c) Corrected Line 4: "Therefore $\triangle ABC \sim \triangle DEF$ by AAA (or AA), but they are NOT necessarily congruent." To upgrade to congruence, we would need to know that at least one pair of corresponding sides is equal (e.g. $AB = DE$). With AA + a matching side, the test becomes ASA.
3 — Open-ended challenge: the ambiguous case (sample solution)
(i) Sketch: Draw segment $AB$ of length $8$. From $A$, draw a ray at $30°$ above $AB$ — this is where $C$ might lie. From $B$, swing an arc of radius $5$ cm. The arc crosses the ray at two points, giving two valid triangles with $AB = 8$, $BC = 5$, $\angle A = 30°$.
(ii) Why two: The $30°$ angle at $A$ fixes the direction of $AC$, but the side $BC = 5$ is "swung" from $B$ — it can meet $AC$ at a near point (forming an acute triangle) or a far point (forming an obtuse triangle). Both have $AB = 8$, $BC = 5$, $\angle A = 30°$, but $AC$ (and hence the triangles) are different.
(iii) Why valid tests work: In SAS, the angle is between the two sides, so once both sides and the angle are fixed, the third side is forced and only one triangle exists. RHS works for right-angled triangles because Pythagoras pins down the third side: $\text{leg}^2 = \text{hyp}^2 - \text{other leg}^2$ has only one positive solution.
Marking: 1 for the sketch idea (two distinct triangles); 1 for an explanation of why two triangles are possible; 1 for connecting to SAS (included angle); 1 for the RHS / Pythagoras reasoning.