GEOM-5.2

Triangle Congruence: SSS, SAS, ASA & AAS

Learn how SSS, SAS, ASA, and AAS prove two triangles congruent — plus clear counterexamples showing why AAA and SSA never work.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on Triangle Congruence: SSS, SAS, ASA & AAS, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

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What this lesson covers

Six pairs of parts — three sides and three angles — have to match for two triangles to be congruent. Checking all six every time would be exhausting, so geometry gives you shortcuts: if the right three pairs match, the other three are forced to match too. SSS, SAS, ASA, and AAS are those shortcuts.

But not every set of three works. AAA and SSA look just as reasonable on paper, and they fail. In this lesson you will learn exactly what each valid criterion requires, how to spot hidden information in a diagram (shared sides, vertical angles, parallel-line angles), and how to build the counterexamples that show why AAA and SSA cannot be trusted. Getting comfortable here pays off immediately, because the next lesson uses these criteria as the engine inside two-column proofs.

Congruence and Correspondence

Two triangles are congruent when there is a correspondence between their vertices such that all three pairs of corresponding sides and all three pairs of corresponding angles are congruent. Equivalently — and this is the definition your course builds on — two figures are congruent when a sequence of rigid motions (translations, rotations, reflections) carries one exactly onto the other. Rigid motions preserve distance and angle measure, which is why matching parts is the same thing as being able to slide, turn, or flip one triangle onto the other.

The correspondence is part of the statement, not an afterthought. Writing △ABC≅△DEF\triangle ABC \cong \triangle DEF claims six specific facts: AB‾≅DE‾\overline{AB} \cong \overline{DE}, BC‾≅EF‾\overline{BC} \cong \overline{EF}, CA‾≅FD‾\overline{CA} \cong \overline{FD}, ∠A≅∠D\angle A \cong \angle D, ∠B≅∠E\angle B \cong \angle E, and ∠C≅∠F\angle C \cong \angle F. Vertices are listed in matching order. If instead the correct pairing sends A→EA \to E, then △ABC≅△DEF\triangle ABC \cong \triangle DEF is simply false even though the two triangles may genuinely be congruent under a different order.

This is where a lot of homework goes wrong: the geometry is right but the letters are scrambled. Before writing any congruence statement, mark the diagram — tick marks on equal sides, arcs on equal angles — and identify which vertex of the first triangle sits opposite which side, then find its partner. A useful check: the angle named first in each triangle must be the pair you actually proved congruent, and the side between the first two letters of each name must be a matching pair as well.

The Four Valid Criteria

Each criterion names three pairs of parts that are enough to force congruence. The word included is doing heavy lifting.
CriterionWhat must be congruentThe restriction that matters
SSSall three pairs of sidesnone — sides alone fix the shape
SAStwo pairs of sides and the pair of angles between themthe angle must be included between those two sides
ASAtwo pairs of angles and the pair of sides between themthe side must be included between those two angles
AAStwo pairs of angles and a pair of non-included sidesthe side must be in the same position relative to both angles
SSS works because three side lengths determine a rigid frame; there is no way to flex a triangle without changing a side. SAS works because once two sides are fixed and the angle between them is fixed, the third vertex position — and therefore the third side — is completely determined. ASA works because the side between two known angles fixes the scale, and the two rays drawn from its endpoints meet at exactly one point.

AAS is really ASA in disguise. If two angles match, the third angles match too, by the Triangle Angle Sum. So a non-included side becomes an included side for a different pair of angles, and ASA applies. That is why AAS is valid while AAA is not: AAS carries one length, and AAA carries none.

When you cite AAS, be careful that the side sits in the same relative spot in both triangles — for instance, opposite the first named angle in each. A side opposite ∠P\angle P in one triangle paired with a side adjacent to the matching angle in the other proves nothing.

Why AAA and SSA Fail

AAA fails because it controls shape but not size. A 33–44–55 right triangle and a 66–88–1010 right triangle have exactly the same three angle measures, yet one is twice as large as the other. No rigid motion can map the small one onto the big one, because rigid motions preserve distance. AAA guarantees similarity, not congruence — a distinction the similarity unit builds on directly.

SSA fails because two sides and a non-included angle can close up in two different ways. Suppose ∠A=30∘\angle A = 30^\circ, AB=8AB = 8, and BC=5BC = 5, where BCBC is opposite the given angle. Put the 30∘30^\circ angle at AA, lay off AB=8AB = 8, then swing a compass of radius 55 from BB. That arc crosses the other ray of the angle at two different points, producing two genuinely different triangles: one with an acute angle at CC (about 53∘53^\circ) and one with an obtuse angle at CC (about 127∘127^\circ). Same three given parts, non-congruent triangles. This is the ambiguous case.

The swinging-arc picture also explains the exception. If the given angle is a right angle and the side opposite it is the hypotenuse, the arc can only meet the other ray once, so HL (Hypotenuse–Leg) for right triangles is valid even though it looks like SSA. HL is a special case, not permission to use SSA generally.

A common wrong move on homework is labeling a setup SSA and calling the triangles congruent "because the parts match." Matching parts are necessary but not sufficient — the arrangement decides.

Finding the Hidden Third Pair

Most problems hand you two pairs of congruent parts openly and hide the third in the picture. Training your eye for these is the real skill.
What you seeWhat you may concludeReason
Two triangles share a sidethat side is congruent to itselfReflexive Property
Two segments crossthe opposite angles are congruentVertical Angles Theorem
Parallel marks plus a transversalalternate interior or corresponding angles congruentparallel line angle relationships
A midpoint marked on a segmentthe two halves are congruentdefinition of midpoint
A segment marked as an angle bisectorthe two angles are congruentdefinition of angle bisector
A right-angle box in each trianglethose angles are congruentall right angles are congruent
A reliable routine: first copy every given onto the diagram with tick marks and arcs; second, scan for the hidden pair from the table above; third, count what you have and check the arrangement — are the sides adjacent to the marked angle, or is the angle floating off to the side? Only then name the criterion.

Two traps show up repeatedly. First, students count a shared side once instead of recognizing it as a legitimate congruent pair in both triangles. Second, students see three marked pairs, assume that is automatically enough, and skip the arrangement check — which is exactly how SSA gets mistaken for SAS. Ask specifically: is the marked angle between the two marked sides? If not, you do not have SAS.

Deciding Which Criterion Applies

Once the diagram is fully marked, sort what you have by counting sides and angles.

Three sides and no angles gives SSS. Two sides and one angle splits in two: if the angle is between the sides, SAS proves congruence; if it is not, you have SSA and the triangles may or may not be congruent — unless the angle is right and the side opposite it is the hypotenuse, which is HL. Two angles and one side also splits in two, but both branches work: the side between the angles gives ASA, and a side outside gives AAS. Three angles and no sides gives AAA, which shows only that the triangles have the same shape.

When a problem asks "is there enough information?", the honest answer is sometimes no. If you are given two sides and an angle in a non-included position with no right angle, write that the triangles cannot be proven congruent and, ideally, sketch the two possible triangles. A correct "not enough information" with justification is a complete answer, while a confident but false SSA claim is simply wrong.

One more habit worth building now: after you name the criterion, immediately write the congruence statement with matching vertex order. In the next lesson, CPCTC lets you conclude that any remaining pair of parts is congruent — but only if the correspondence in your statement is correct. A sloppy vertex order there produces a conclusion about the wrong pair of parts, and everything after it collapses.

Key terms

Congruent triangles.
Triangles whose three pairs of corresponding sides and three pairs of corresponding angles are all congruent; equivalently, one can be mapped onto the other by a sequence of rigid motions.
Correspondence.
The specific pairing of vertices between two triangles, shown by the order of letters in a statement such as △ABC≅△DEF\triangle ABC \cong \triangle DEF.
Included angle.
The angle formed by two named sides of a triangle — the angle at the vertex where those two sides meet. Required for SAS.
Included side.
The side that lies between two named angles of a triangle, connecting their vertices. Required for ASA.
AAS.
Angle–Angle–Side: two pairs of congruent angles plus a pair of congruent non-included sides in matching positions. Valid because the third angles must also match.
Ambiguous case (SSA).
A setup with two sides and a non-included angle, in which the third vertex can often land in two different places, producing two non-congruent triangles.
HL.
Hypotenuse–Leg: for right triangles only, congruent hypotenuses and one pair of congruent legs prove congruence; the sole valid SSA-shaped criterion.
Reflexive Property of Congruence.
Any segment or angle is congruent to itself; used to claim a shared side or shared angle as one of the three needed pairs.

Worked example

In quadrilateral ABCDABCD, AB‾≅CD‾\overline{AB} \cong \overline{CD} and AB‾∥CD‾\overline{AB} \parallel \overline{CD}. Diagonal AC‾\overline{AC} is drawn, forming △ABC\triangle ABC and △CDA\triangle CDA. Determine whether the two triangles are congruent, name the criterion, and write the congruence statement.
Start by listing what is given and marking it. Given pair one: AB‾≅CD‾\overline{AB} \cong \overline{CD} — one pair of sides. Given pair two: AB‾∥CD‾\overline{AB} \parallel \overline{CD} — this is not a congruence yet, but parallel lines with a transversal produce congruent angles.

Identify the transversal. The diagonal AC‾\overline{AC} cuts across both parallel segments, so ∠BAC\angle BAC and ∠DCA\angle DCA are alternate interior angles. Therefore ∠BAC≅∠DCA\angle BAC \cong \angle DCA. That is pair two.

Now hunt for the hidden third pair. The two triangles share the diagonal, so AC‾≅CA‾\overline{AC} \cong \overline{CA} by the Reflexive Property. That is pair three.

Check the arrangement before naming anything. In △ABC\triangle ABC, the marked angle is ∠BAC\angle BAC, at vertex AA. The two marked sides are AB‾\overline{AB} and AC‾\overline{AC} — both of them meet at AA. So the angle is included between the two sides. In △CDA\triangle CDA, the marked angle is ∠DCA\angle DCA at vertex CC, and the marked sides CD‾\overline{CD} and CA‾\overline{CA} both meet at CC. Same arrangement.

Side–included angle–side in both triangles means the criterion is SAS.

Finally, write the correspondence so matching parts line up: A→CA \to C (the marked angles), B→DB \to D, and C→AC \to A. The statement is △ABC≅△CDA\triangle ABC \cong \triangle CDA. Notice that AB‾\overline{AB} and CD‾\overline{CD} occupy the first two letters in each name, matching the given side pair — a quick way to confirm the order is right.

Practice questions

In △PQR\triangle PQR and △STU\triangle STU, ∠P≅∠S\angle P \cong \angle S, ∠Q≅∠T\angle Q \cong \angle T, and QR‾≅TU‾\overline{QR} \cong \overline{TU}. Which criterion, if any, proves the triangles congruent?
  1. ASA, because two angles and a side are congruent
  2. AAS, because the congruent sides are not between the two pairs of congruent angles
  3. SAS, because an angle sits between two known parts
  4. No criterion applies; the information is insufficient

Answer: AAS, because the congruent sides are not between the two pairs of congruent angles

The two marked angles are at PP and QQ, so the included side for those angles would be PQ‾\overline{PQ}. The given side is QR‾\overline{QR}, which touches ∠Q\angle Q but not ∠P\angle P — it is non-included, which is AAS, not ASA. Check the matching position in the other triangle: the marked angles are at SS and TT, and TU‾\overline{TU} touches ∠T\angle T but not ∠S\angle S. Same position, so AAS applies and △PQR≅△STU\triangle PQR \cong \triangle STU. AAS is legitimate because the third angles, ∠R\angle R and ∠U\angle U, must also be congruent by the Triangle Angle Sum, which converts the situation into ASA.
A classmate claims that if ∠A≅∠D\angle A \cong \angle D, AB‾≅DE‾\overline{AB} \cong \overline{DE}, and BC‾≅EF‾\overline{BC} \cong \overline{EF}, then △ABC≅△DEF\triangle ABC \cong \triangle DEF. Explain why this reasoning is not valid, and describe a specific pair of triangles that shows it fails.

Answer: The given angle is not included between the two given sides, so this is SSA, which does not guarantee congruence. Counterexample: let ∠A=30∘\angle A = 30^\circ, AB=8AB = 8, and BC=5BC = 5. Swinging an arc of radius 5 from BB meets the other ray of the angle at two points, giving one triangle with ∠C≈53∘\angle C \approx 53^\circ and another with ∠C≈127∘\angle C \approx 127^\circ. Both satisfy all three given conditions but are not congruent.

The key is to test the arrangement, not just count parts. In △ABC\triangle ABC, sides AB‾\overline{AB} and BC‾\overline{BC} meet at BB, so the included angle would be ∠B\angle B — but the given angle is at AA, opposite BC‾\overline{BC}. That is the SSA pattern. A construction argument is the cleanest justification: fix the angle at AA, draw the side of length 8, then the third vertex must lie both on the far ray and at distance 5 from BB; a circle can cross a line twice, so two triangles satisfy the conditions. The claim would become valid only if ∠A\angle A were a right angle, in which case the arc meets the ray once and HL applies.
Segments JM‾\overline{JM} and KL‾\overline{KL} intersect at point NN, with NN the midpoint of both segments. Name a criterion that proves △JNK≅△MNL\triangle JNK \cong \triangle MNL and justify each pair you use.

Answer: SAS. Since NN is the midpoint of JM‾\overline{JM}, JN‾≅MN‾\overline{JN} \cong \overline{MN}; since NN is the midpoint of KL‾\overline{KL}, KN‾≅LN‾\overline{KN} \cong \overline{LN}; and ∠JNK≅∠MNL\angle JNK \cong \angle MNL because they are vertical angles. The angle is included between the two sides in each triangle, so SAS applies.

Two hidden facts do all the work here. The midpoint marks give two pairs of congruent segments straight from the definition of midpoint. The crossing segments create vertical angles at NN, which are always congruent. Then verify inclusion: in △JNK\triangle JNK, the sides JN‾\overline{JN} and KN‾\overline{KN} both meet at NN, and ∠JNK\angle JNK is exactly the angle at NN between them. The same holds in △MNL\triangle MNL. Students who list the parts as "side, side, angle" without checking where the angle sits sometimes reject this as SSA — always locate the angle's vertex relative to the two sides before deciding.

FAQ

Is AAS the same thing as SAA?
Yes, they refer to the same criterion; the letters just read the parts in the opposite order around the triangle. What matters is that two angles and a non-included side are congruent, and that the side occupies the same position relative to those angles in both triangles. Use whichever abbreviation your class uses, and stay consistent within a proof.
Why is HL allowed if SSA is not?
HL has the same shape as SSA — two sides and a non-included angle — but the given angle is a right angle and the given side opposite it is the hypotenuse, the longest side. In the compass-swing picture, an arc of that length from the far vertex can only cross the other ray once, so no second triangle exists. Remove the right angle and the ambiguity comes right back, which is why HL is restricted to right triangles.
Does the order of letters in a congruence statement really matter?
Yes. △ABC≅△DEF\triangle ABC \cong \triangle DEF asserts six specific matchings, including ∠A≅∠D\angle A \cong \angle D and BC‾≅EF‾\overline{BC} \cong \overline{EF}. If the actual correspondence pairs AA with EE, the statement is false even though the triangles are congruent. This becomes critical in the next lesson, where CPCTC pulls conclusions directly out of the order you wrote.
What should I write if none of the four criteria apply?
State clearly that the triangles cannot be proven congruent from the given information, name the pattern you see (usually SSA or AAA), and back it up. For AAA, mention that the triangles are similar but could be different sizes; for SSA, sketch or describe the two possible triangles. A justified "not enough information" is a complete answer, not a dodge.

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