Go deeper into the four tests. Learn to write formal proofs that examiners love -- clear, logical, and impossible to argue with.
Today's hook: Two triangles share three matching parts. Is that always enough to prove they are identical? What if the matching angle is not between the matching sides -- does that change anything?
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Triangle $ABC$ has $AB = 5$ cm, $BC = 6$ cm, and $\angle B = 50°$. Triangle $DEF$ has $DE = 5$ cm, $EF = 6$ cm, and $\angle E = 50°$. Are the triangles definitely congruent? What if $\angle A = 50°$ instead of $\angle B$ -- would they still be congruent? Explain.
Record your answer in your workbook.
1
The Big Idea
+5 XP to read
A formal congruence proof is not guesswork -- it is a logical chain. You name the triangles, list the matching parts with reasons, state the test, and conclude. Every step must be justified. Get this structure right and you will earn full marks in any exam.
The four tests -- SSS, SAS, AAS, RHS -- are the only valid ways to prove congruence. SSA is never valid except in the special RHS case. The key skill is not memorising the tests but recognising which test fits the given information and writing the proof clearly.
given $\to$ test $\to$ conclude
Name the triangles first
Always begin: "In triangles ABC and DEF..."
Give reasons
Every matching part needs a justification.
End with the test
Finish: "Therefore ... (SAS) in brackets."
2
What You'll Master
objectives
Know
The four congruence tests: SSS, SAS, AAS, RHS
Why matching three corresponding parts proves identity
Understand
Why the included angle is essential for SAS
Why SSA is not valid (the ambiguous case)
The difference between similar and congruent triangles
Can Do
Identify the correct test for any pair of triangles
Write formal congruence proofs with clear reasoning
Use congruence to find unknown sides and angles
3
Words You Need
vocabulary
SSSSide-Side-Side: all three corresponding sides are equal.
SASSide-Angle-Side: two corresponding sides and the included angle are equal.
AASAngle-Angle-Side: two corresponding angles and a non-included side are equal.
RHSRight angle-Hypotenuse-Side: right triangles with matching hypotenuse and one leg.
CPCTCCorresponding Parts of Congruent Triangles are Congruent. Used after proving congruence.
Ambiguous caseWhen SSA measurements can produce two different triangles.
4
Spot the Trap
heads-up
Wrong: SSA (Side-Side-Angle) is a valid congruence test.
Right: SSA is not a valid test. It is called the ambiguous case because two different triangles can share the same SSA measurements.
Wrong: Similar triangles are congruent.
Right: Similar triangles have the same shape but not necessarily the same size. Congruent triangles have both the same shape and the same size.
5
SSS and SAS
+5 XP
SSS is the most direct test: match all three sides and the triangles are rigid. SAS is equally powerful but demands the angle be included between the two sides. This inclusion condition is what makes SAS work and SSA fail.
In SAS, the included angle fixes the distance between the two side endpoints. The angle is the hinge that determines the triangle's shape. Move the angle outside the sides and you get SSA -- where the third side can swing to two different positions.
SAS = rigid | SSA = ambiguous
Check inclusion
For SAS, the angle must sit between the two sides.
SSS needs all three
Any three matching sides guarantees congruence.
Label diagrams
Mark equal sides and angles before choosing.
6
AAS and RHS
+5 XP
AAS works because two angles fix the shape, and one side fixes the size. Since the third angle is automatically known (angles in a triangle sum to 180°), AAS is really "two angles and any corresponding side." RHS is the special case where SSA works -- because the right angle removes the ambiguity.
In AAS, once two angles are fixed, the third follows from $180° - a - b$. The side then fixes the scale. RHS is essentially SSS in disguise -- once you know the hypotenuse and one leg in a right triangle, Pythagoras gives you the third side.
$a = \sqrt{h^2 - b^2}$ by Pythagoras
AAS = shape + size
Two angles fix shape. One side fixes scale.
RHS = right triangles only
Needs 90°, hypotenuse, and one leg.
Third angle is free
In AAS, the third angle comes from 180°.
7
The Ambiguous Case
+5 XP
SSA fails because the same two sides and non-included angle can produce two different triangles. Imagine holding two sticks of fixed length at one end, with a fixed angle at the other end -- the third vertex can swing to two positions. This is why SSA is never accepted as a congruence test.
The only exception is RHS. In a right triangle, the hypotenuse and one leg fix everything because Pythagoras determines the third side. The right angle constraint pins the triangle down -- there is no second position for the third vertex.
SSA $\neq$ congruence
Check angle position
If the angle is not between the sides, it is SSA.
RHS is the only exception
The right angle fixes the third side via Pythagoras.
Draw to verify
Sketch both possibilities to see the ambiguity.
Watch Me Solve It · 3 examples
Watch Me Solve It · Identifying the test
+15 XP per step
Q1
PROBLEM
Triangle $ABC$ has $AB = 5$ cm, $BC = 6$ cm, $AC = 7$ cm. Triangle $DEF$ has $DE = 5$ cm, $EF = 6$ cm, $DF = 7$ cm. Which test proves congruence?
1
List the matching parts
$AB = DE = 5$ cm, $BC = EF = 6$ cm, $AC = DF = 7$ cm
All three pairs of corresponding sides are given as equal.
2
Count the matching parts
Three sides match -- this is the SSS condition
3
Write the conclusion
$\triangle ABC \equiv \triangle DEF$ (SSS)
Vertices match in the order given: $A \leftrightarrow D$, $B \leftrightarrow E$, $C \leftrightarrow F$.
Nice work -- XP earned
Answer$\triangle ABC \equiv \triangle DEF$ (SSS)
Watch Me Solve It · Formal SAS proof
+15 XP per step
Q2
PROBLEM
Prove that $\triangle ABC \equiv \triangle DEF$, given that $AB = DE$, $BC = EF$, and $\angle B = \angle E$.
1
Name the triangles
In triangles $ABC$ and $DEF$:
Always begin a formal proof by naming both triangles.
2
List the given information with reasons
$AB = DE$ (given)
$\angle B = \angle E$ (given)
$BC = EF$ (given)
3
Identify the test
Two sides and the included angle are equal → SAS
The angle $\angle B$ is between sides $AB$ and $BC$.
The symbol $\therefore$ means "therefore." The test name in brackets is required.
Nice work -- XP earned
Answer$\triangle ABC \equiv \triangle DEF$ (SAS)
Watch Me Solve It · Finding unknowns
+15 XP per step
Q3
PROBLEM
$ABCD$ is a parallelogram. Prove $\triangle ABC \equiv \triangle CDA$ and hence find $\angle BAC$ if $\angle DCA = 35°$.
1
Name the triangles and list matching sides
In triangles $ABC$ and $CDA$:
$AB = CD$ (opposite sides of parallelogram)
$BC = DA$ (opposite sides of parallelogram)
$AC = CA$ (common side)
2
State the congruence
$\triangle ABC \equiv \triangle CDA$ (SSS)
All three pairs of corresponding sides are equal.
3
Use CPCTC to find the unknown angle
$\angle BAC = \angle DCA$ (corresponding angles of congruent triangles)
$\angle BAC$ corresponds to $\angle DCA$ because $A \leftrightarrow C$ and $C \leftrightarrow A$ in the congruence statement.
4
Substitute the known value
$\angle BAC = 35°$
Nice work -- XP earned
Answer$\angle BAC = 35°$
9
Common Pitfalls
heads-up
Using SSA as a valid test
Selecting or writing SSA in a proof. Two different triangles can share the same two sides and non-included angle, so SSA never proves congruence.
Fix: check if the angle is included between the two sides. If not, it is SSA and invalid. The only exception is RHS for right triangles.
Confusing similar with congruent
Claiming similar triangles are congruent. Similar triangles have equal angles but sides in proportion -- they can be different sizes. Congruent triangles have equal sides and equal angles.
Fix: similar = same shape. congruent = same shape AND same size.
Missing reasons in proofs
Writing "$AB = DE$" without "(given)" or "(opposite sides of parallelogram)." In formal proofs, every statement needs a reason.
Fix: get into the habit of writing the reason in brackets after every line.
Copy Into Your Books
SSS
Three sides equal
Rigid triangle frame
No angles needed
SAS
Two sides + included angle
Angle must be between sides
Most common test in proofs
AAS and RHS
AAS: two angles + side
RHS: right triangle only
Both fix shape and size
Proof Structure
Name both triangles
List matching parts + reasons
State test in brackets
Use CPCTC for unknowns
How are you completing this lesson?
Brain Trainer · 4 problems
D
Brain Trainer · Test identification
4 problems
Identify the congruence test for each pair, or state "not possible."
1 $\triangle ABC$: $AB = 8$ cm, $\angle A = 50°$, $AC = 10$ cm. $\triangle DEF$: $DE = 8$ cm, $\angle D = 50°$, $DF = 10$ cm.
SAS. Two sides and the included angle are equal. $\angle A$ is between $AB$ and $AC$; $\angle D$ is between $DE$ and $DF$.SAS
2 $\triangle ABC$: $\angle A = 40°$, $\angle B = 60°$, $AB = 9$ cm. $\triangle DEF$: $\angle D = 40°$, $\angle E = 60°$, $DE = 9$ cm.
AAS. Two angles and a corresponding side. Since two angles are equal, the third must also be equal ($80°$).AAS
3 $\triangle ABC$: $AB = 6$ cm, $BC = 8$ cm, $\angle A = 30°$. $\triangle DEF$: $DE = 6$ cm, $EF = 8$ cm, $\angle D = 30°$. The angle is not included.
Not possible. This is SSA -- two sides and a non-included angle. Two different triangles could satisfy these conditions.Not possible
4 Two right triangles have hypotenuse 15 cm and one leg 9 cm.
RHS. Right angle, hypotenuse, and one side. The other leg is fixed by Pythagoras ($\sqrt{15^2 - 9^2} = 12$ cm), making this equivalent to SSS.RHS
Complete in your workbook.
Multiple Choice · 5 questions
MC1
All three sides
+10 XP
Which test proves congruence when all three sides are equal?
Correct -- SSS (Side-Side-Side) matches all three corresponding sides.
Three equal sides means SSS. SAS needs an included angle, AAS needs angles, RHS needs a right triangle.
Explanation: SSS is the test where all three pairs of corresponding sides are equal. It is the most direct test and guarantees the triangles are identical.
MC2
The included angle
+10 XP
For SAS congruence, the angle must be:
Correct -- SAS requires the angle to be included between the two known sides.
SAS specifically means Side-Angle-Side where the angle is between the two sides.
Explanation: The "A" in SAS stands for the included angle -- the angle formed by the two sides. If the angle is not between the sides, it becomes SSA, which is not a valid test.
MC3
RHS restriction
+10 XP
RHS congruence applies only to:
Correct -- RHS (Right angle-Hypotenuse-Side) is specifically for right-angled triangles.
The R in RHS stands for Right angle. This test only applies to right-angled triangles.
Explanation: RHS requires a right angle, a matching hypotenuse, and one matching leg. It is the only case where SSA works, because the right angle constraint fixes the third side via Pythagoras.
MC4
AAS parts
+10 XP
If $\triangle ABC \cong \triangle DEF$ by AAS, which parts match?
Correct -- AAS is Angle-Angle-Side: two angles and a non-included side.
AAS specifically uses two angles and a side that is not between them.
Explanation: AAS stands for Angle-Angle-Side. The side is not included between the two angles. Since two angles fix the shape and one side fixes the size, this is sufficient to prove congruence.
MC5
Congruent definition
+10 XP
Congruent triangles have:
Correct -- congruent means identical in every way: same shape and same size.
Same angles only means similar, not necessarily congruent. Congruent requires equal sides too.
Explanation: Congruent triangles are identical -- every corresponding side and angle is equal. Similar triangles have the same shape (equal angles) but may be different sizes. Same area or perimeter alone does not guarantee congruence.
Short Answer · 3 questions
Q6
SSA analysis
+15 XP
Q6
SHORT ANSWER
In triangles $ABC$ and $DEF$, $AB = DE$, $AC = DF$, and $\angle B = \angle E$. Can you prove that the triangles are congruent? Explain why or why not.
Write your working in your book.
No. This is SSA, not SAS.
The given information is: side $AB$, side $AC$, and angle $\angle B$. The angle $\angle B$ is not included between sides $AB$ and $AC$ -- it is at vertex $B$, between sides $AB$ and $BC$.
Since we have two sides and a non-included angle, this is the ambiguous case. Two different triangles could satisfy these conditions. Therefore we cannot prove congruence.
Marking guidance: 1 mark for correct claim, 1 mark for identifying SSA, 1 mark for explaining the ambiguous case.
Q7
Parallelogram proof
+15 XP
Q7
SHORT ANSWER
Prove that the two triangles formed by the diagonal of a parallelogram are congruent.
Write your working in your book.
Let $ABCD$ be a parallelogram and let $AC$ be a diagonal.
Marking guidance: 1 mark for naming triangles, 1 mark for two correct reasons, 1 mark for conclusion with test.
Q8
The ambiguous case
+15 XP
Q8
SHORT ANSWER
Explain why SSA is not a valid congruence test. Draw a diagram to support your answer.
Write your working in your book.
SSA is not valid because the same two sides and non-included angle can produce two different triangles.
Consider a triangle with sides 5 cm and 7 cm, and a non-included angle of 30°. Draw the 7 cm side as a base. At one end, construct the 30° angle. From the other end, draw an arc of radius 5 cm. This arc can intersect the ray from the 30° angle at two different points, creating two different triangles that both satisfy the given SSA conditions.
Since two different triangles are possible, SSA does not uniquely determine a triangle and cannot be used as a congruence test.
Marking guidance: 1 mark for explanation, 1 mark for diagram description, 1 mark for clear conclusion.
S
Stretch Challenge · Two proofs in one
+20 XP
S
STRETCH
$ABCD$ is a kite with $AB = AD$ and $CB = CD$. The diagonals $AC$ and $BD$ intersect at $E$. (a) Prove that $\triangle ABC \equiv \triangle ADC$. (b) Prove that $AC$ is the perpendicular bisector of $BD$. (c) Hence explain why a kite has exactly one line of symmetry.
Record in your book -- full marks require clear working.
(b) From congruence, $\angle BAE = \angle DAE$ and $\angle BCE = \angle DCE$. In triangles $ABE$ and $ADE$: $AB = AD$ (given), $\angle BAE = \angle DAE$ (from part a), $AE = AE$ (common). Therefore $\triangle ABE \equiv \triangle ADE$ (SAS), so $BE = DE$ and $\angle AEB = \angle AED = 90°$. Thus $AC$ bisects $BD$ at right angles.
(c) Since $AC$ is the perpendicular bisector of $BD$, reflecting across $AC$ maps $B$ to $D$ and keeps $A$ and $C$ fixed. This reflection symmetry means $AC$ is the only line of symmetry.
R
Quick Review
SSS
Three sides equal
SAS
Two sides, included angle
AAS
Two angles, side
RHS
Right triangle special case
SSA fails
Ambiguous case
Proof steps
Name, list, test, conclude
Interactive: Angle Chase
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Consolidation Game -- Doodle Jump Quiz
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