Formal Congruence Proofs
Congruent triangles are identical in every measurement. Proving it takes exactly three facts, chosen to fit one of four tests, and written with the vertices in matching order.
Sketch a triangle with sides $3$ cm, $4$ cm and $5$ cm. Now try to sketch a genuinely different triangle with the same three side lengths. Can you? Now sketch a triangle with angles $50°$, $60°$ and $70°$, and try to draw a different one with those same three angles. What is the difference between the two situations?
Two triangles are congruent when every matching side and every matching angle is equal. You never have to check all six: three well-chosen facts force the other three. There are exactly four such combinations, and a proof consists of establishing three facts, naming the test, and writing the congruence with the vertices in matching order.
$$\text{SSS} \qquad \text{SAS} \qquad \text{AAS} \qquad \text{RHS}$$
The commonest third fact is a shared side. When two triangles sit either side of a diagonal, that diagonal belongs to both, and "$AC = AC$, common" is a legitimate line of proof. Looking for the shared element before hunting for anything else will solve a large proportion of congruence questions.
Know
- The four congruence tests: SSS, SAS, AAS and RHS
- That AAA and SSA are not tests, and why
- That the congruence statement records which vertex matches which
Understand
- Why three well-chosen facts force the remaining three
- Why the angle in SAS must be the included one
Can Do
- Set out a formal congruence proof with a reason on every line
- Choose the appropriate test from the facts available
- Draw further conclusions from a congruence, citing matching sides or angles
Each test is a set of three facts sufficient to force congruence.
| Test | What you need | Watch out for |
|---|---|---|
| SSS | all three pairs of sides equal | nothing; the easiest to justify |
| SAS | two pairs of sides, and the angle BETWEEN them | an angle not between the two sides does not count |
| AAS | two pairs of angles and one pair of matching sides | the side must be in the same position in both |
| RHS | right angle, hypotenuse, one other side | only for right-angled triangles |
AAS is worth a note. Once two angles are known the third is fixed, because the angle sum is $180°$. So AAS really says "the same three angles, plus one actual length to set the size", and the position of that length must match.
AAA fails because angles fix shape, not size. A triangle with angles $50°$, $60°$ and $70°$ can be drawn a centimetre across or a kilometre across. Every such triangle has the same shape, so the triangles are similar, which is the subject of Lesson 3, but they need not be congruent.
SSA fails because the angle is not included. Given two sides and an angle NOT between them, two genuinely different triangles can often be built: the third side can swing to meet the base in two places. This is the ambiguous case, and it is exactly why SAS insists on the included angle.
RHS is the one exception that looks like SSA, and it works because a right angle removes the ambiguity: with the right angle and the hypotenuse fixed, Pythagoras determines the remaining side uniquely, so only one triangle can be built.
Questions usually give you two facts and expect you to find the third. Three sources supply it almost every time.
A common side. When two triangles share an edge, that edge is equal to itself. Write "$AC = AC$ (common)". This is the most frequent third fact in the whole topic.
Parallel lines. If the diagram has parallel arrows, alternate or corresponding angles give you an equal pair for free.
A definition unpacked. "$M$ is the midpoint of $QR$" is one word that yields $QM = MR$. "$AD$ bisects $\angle BAC$" yields $\angle BAD = \angle CAD$. Read the question for these before deciding you are short of information.
Vertically opposite angles are a fourth source, appearing whenever two lines cross between the triangles.
Once the test is named, the congruence statement must record which vertex matches which:
$$\triangle ABD \equiv \triangle ACD \quad \text{(SAS)}$$
That single line asserts $A \leftrightarrow A$, $B \leftrightarrow C$ and $D \leftrightarrow D$. Every further conclusion is then read straight off it:
$$BD = CD \quad \text{(matching sides of congruent triangles)}$$
$$\angle ABD = \angle ACD \quad \text{(matching angles of congruent triangles)}$$
Get the order wrong and every conclusion drawn from it is wrong, even though the triangles genuinely are congruent. Before writing the statement, mark the matching vertices on the diagram and read the letters off the marks, not off the order the question happened to use.
Watch Me Solve It · 3 examples
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1Draw the diagonal and list the given facts$AB = CD \quad \text{(given)}$$AD = CB \quad \text{(given)}$Joining $A$ to $C$ creates two triangles that share that diagonal.
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2Supply the third fact$AC = AC \quad \text{(common side)}$The diagonal belongs to both triangles, so it equals itself.
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3Name the test, in matching order$\triangle ABC \equiv \triangle CDA \quad \text{(SSS)}$$A$ matches $C$, $B$ matches $D$, $C$ matches $A$: read from the equal sides, not the alphabet.
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4Draw the conclusion$\angle B = \angle D \quad \text{(matching angles of congruent triangles)}$$B$ and $D$ are matching vertices in the statement above.
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1Use the parallel lines$\angle BAM = \angle DCM \quad \text{(alternate angles, } AB \parallel DC)$$\angle ABM = \angle CDM \quad \text{(alternate angles, } AB \parallel DC)$Parallel arrows on a diagram are given information and hand you two angles.
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2Add the given side$AB = DC \quad \text{(given)}$
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3Identify the test$\text{two angles and the side between them: AAS}$The side is in the same position relative to the two angles in both triangles, which is the condition AAS requires.
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4State the congruence$\triangle ABM \equiv \triangle CDM \quad \text{(AAS)}$$A$ matches $C$, $B$ matches $D$, $M$ matches $M$.
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1Identify the hypotenuses$PR \text{ is opposite } \angle Q, \quad SU \text{ is opposite } \angle T$The hypotenuse is always the side opposite the right angle.
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2List the three facts$\angle Q = \angle T = 90°$$PR = SU \quad \text{(hypotenuses)}$$PQ = ST \quad \text{(one other side)}$
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3Name the test$\triangle PQR \equiv \triangle STU \quad \text{(RHS)}$
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4Explain why the right angle matters$QR^2 = PR^2 - PQ^2$Pythagoras fixes the third side uniquely, so only one triangle can be built. Without a right angle the same three facts are SSA and the third side can take two different values.
Brain Trainer · 4 problems
Four quick problems. Work each one, then reveal the answer.
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1 Two triangles have all three pairs of sides equal. Which test?
Three sides is the most direct test of all.SSS -
2 Two triangles have angles $40°$ and $70°$ equal, and the sides between those angles equal. Which test?
Two angles and a matching side.AAS -
3 Two triangles have three pairs of equal angles. Are they congruent?
Angles fix the shape but not the size.No, only similar -
4 Two triangles share a side. What line records this?
A shared side is equal to itself."common side"
Multiple Choice · 5 questions
SAS requires that the equal angle is:
Two triangles have three pairs of equal angles. They must be:
In a proof, the line "$AC = AC$ (common)" is:
From $\triangle ABD \equiv \triangle ACD$, which of these follows?
RHS is a valid test, even though it uses two sides and a non-included angle, because:
Short Answer · 3 questions
(a) Write down the two facts that follow from the word "midpoint".
(b) Identify a third fact available from the diagram, with its reason.
(c) Prove that $\triangle AOB \equiv \triangle COD$, and hence that $AB \parallel CD$.
(a) $AB = PQ$, $BC = QR$, $\angle B = \angle Q$
(b) $AB = PQ$, $BC = QR$, $\angle A = \angle P$
(c) $\angle A = \angle P$, $\angle B = \angle Q$, $\angle C = \angle R$
(d) $\angle B = \angle Q = 90°$, $AC = PR$, $AB = PQ$
"$PQ = SR$ (given). $\angle P = \angle S$ (given). $QR = PS$ (they look equal). So $\triangle PQR \equiv \triangle SRQ$ (SSS), and therefore $\angle Q = \angle R$."
(a) Identify every fault.
(b) Rewrite the proof correctly, assuming instead that $QR$ is a side shared by both triangles.
(a) Explain why these facts are SSA rather than SAS.
(b) Show that two different triangles satisfy them, by finding the two possible values of $\angle C$.
(c) State what extra piece of information would reduce it to one triangle, and justify your answer.
Four tests
SSS, SAS, AAS, RHS
Not tests
AAA gives similarity; SSA is ambiguous
Third fact
Common side, parallel lines, or a definition
Order
Matching vertices license every conclusion
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