What Is the SAS Criterion in Triangles?
The SAS criterion in triangles (Side-Angle-Side) states that two triangles are congruent if two sides of one triangle are equal to two sides of the other, and the angle between those two sides - the included angle - is also equal. When this holds, the triangles match exactly in all three sides and all three angles.
If you deleted the rest of this page, that sentence would still answer the query: two sides plus the angle between them lock a triangle's shape and size completely.
What Does "Included Angle" Mean?
The included angle is the angle formed between the two sides you are comparing - it sits at the vertex where those two sides meet. In triangle $ABC$, the angle included between sides $AB$ and $AC$ is $∠A$. This word "included" is the whole point of SAS: the angle must be the one wedged between the two sides, not one of the other two angles. Get the position of the angle wrong and the rule no longer applies.
What Is the Formal Statement of the SAS Criterion?
For triangles $ABC$ and $DEF$:
$$\text{If } AB = DE,\ ∠A = ∠D,\ \text{and } AC = DF, \text{ then } \triangle ABC \cong \triangle DEF.$$
Key to the statement:
$AB = DE$ - the first pair of sides are equal.
$∠A = ∠D$ - the included angles (each between the two named sides) are equal.
$AC = DF$ - the second pair of sides are equal.
$\cong$ - the symbol for "is congruent to," meaning identical in shape and size.
Once congruence is established, every remaining part matches: $BC = EF$, $∠B = ∠E$, and $∠C = ∠F$. These follow automatically and are often abbreviated CPCT - corresponding parts of congruent triangles. This is one of the core rules of congruence in triangles.
How Do You Prove the SAS Criterion?
SAS can be justified by rigid transformations - slides, turns, and flips that never change size or shape. The idea is to move one triangle exactly onto the other.
Given $\triangle ABC$ and $\triangle DEF$ with $AB = DE$, $∠A = ∠D$, and $AC = DF$:
Translate $\triangle ABC$ so that vertex $A$ lands on vertex $D$.
Rotate it about $D$ so that side $AB$ falls along side $DE$. Because $AB = DE$, point $B$ lands exactly on point $E$.
Since $∠A = ∠D$, side $AC$ now points along the direction of $DF$. Because $AC = DF$, point $C$ lands exactly on point $F$.
With $A$ on $D$, $B$ on $E$, and $C$ on $F$, the two triangles coincide completely - so $\triangle ABC \cong \triangle DEF$.
Because a rigid motion mapped one triangle onto the other with no gaps or overlaps, the triangles must be congruent. This is the modern version of Euclid's original superposition argument.
Why Does SAS Work but SSA Does Not?
This is the most-asked question about SAS, and the answer is about the position of the angle.
SAS works because two sides and the angle between them leave nothing free to change. Fix the angle at a vertex, fix the two edges coming out of it, and the far ends of those edges are pinned - there is only one place the third side can go. One unique triangle results.
SSA fails because the angle is not included - it sits outside the two given sides. With the angle loose at one end, the second side can often swing to two different closing positions, producing two different triangles from the same three measurements. Because two non-congruent triangles can share the same SSA data, SSA is not a valid congruence rule.
The lesson: in SAS the angle is caged between the two sides; in SSA it is free to let the triangle bend.
How Does SAS Compare to the Other Congruence Rules?
SAS is one of a family. Each rule fixes just enough information to pin a triangle:
SSS - three pairs of sides equal. Side-Side-Side congruence needs no angle at all.
ASA - two angles and the included side equal, per the ASA congruence rule.
AAS - two angles and a non-included side equal.
RHS - in right triangles, the right angle, hypotenuse, and one side equal.
SAS is often the most efficient in construction and proof because two lengths and one angle are frequently the easiest measurements to take directly.
Examples of the SAS Criterion in Triangles
The examples move from a direct check to a two-step proof, with one common trap along the way.
Example 1
In triangles $ABC$ and $PQR$, $AB = PQ = 5$ cm, $∠A = ∠P = 60°$, and $AC = PR = 7$ cm. Are the triangles congruent?
Two sides and the included angle match:
$AB = PQ$, $∠A = ∠P$ (included), $AC = PR$.
By SAS, $\triangle ABC \cong \triangle PQR$.
Final answer: yes, congruent by SAS.
Example 2
In triangles $ABC$ and $DEF$, $AB = DE = 6$ cm, $AC = DF = 8$ cm, and $∠B = ∠E = 50°$. A student claims SAS proves them congruent. Are they right?
The tempting path is to say "two sides equal, one angle equal - that's SAS." But look at which angle is given: $∠B$ sits between sides $AB$ and $BC$, and $BC$ is not one of the sides we know. The equal angle is not included between the two equal sides $AB$ and $AC$.
That makes this SSA, not SAS - and SSA does not guarantee congruence. The claim breaks because the angle is in the wrong place.
The correct conclusion: SAS cannot be applied here. The triangles may or may not be congruent; this information is not enough to decide.
Final answer: no - this is SSA, not SAS, so congruence is not proven.
Example 3
Two sides of a triangle are 9 cm and 4 cm with a $90°$ included angle. A second triangle has the same two sides and included angle. Must they be congruent?
Both triangles fix the same two legs and the same $90°$ between them.
By SAS, the triangles are congruent - and here they also happen to be right triangles, so the third sides (the hypotenuses) match too.
Final answer: yes, congruent by SAS.
Example 4
In quadrilateral figure, $O$ is the midpoint of both $AC$ and $BD$. Prove $\triangle AOB \cong \triangle COD$.
Since $O$ is the midpoint of $AC$: $AO = CO$.
Since $O$ is the midpoint of $BD$: $BO = DO$.
The angles $∠AOB$ and $∠COD$ are vertically opposite, so $∠AOB = ∠COD$, and this angle is included between the equal sides.
By SAS, $\triangle AOB \cong \triangle COD$. A common first-instinct error here is to reach for the un-drawn sides $AB$ and $CD$; the vertical angle is the included angle that makes SAS work directly.
Final answer: congruent by SAS, using the vertically opposite included angle.
Example 5
Triangle $ABC$ is isosceles with $AB = AC$. The line $AD$ bisects $∠A$, meeting $BC$ at $D$. Prove $\triangle ABD \cong \triangle ACD$.
$AB = AC$ (given).
$∠BAD = ∠CAD$ (since $AD$ bisects $∠A$) — this is the included angle.
$AD = AD$ (common side).
By SAS, $\triangle ABD \cong \triangle ACD$.
Final answer: congruent by SAS; this also proves the base angles of an isosceles triangle are equal.
Example 6
Two triangles have sides 12 m and 5 m. In the first the included angle is $40°$; in the second it is $70°$. Are they congruent by SAS?
The two sides match, but the included angles differ ($40° \neq 70°$).
SAS requires all three parts - both sides and the included angle - to match. One part fails, so SAS does not apply, and the triangles are not congruent.
Final answer: no - the included angles are different, so SAS fails.
Why Does the SAS Criterion Matter?
"Two lengths and the angle between them are enough to copy a shape exactly." That guarantee is what makes SAS useful far beyond the classroom.
Manufacturing and construction. A part specified by two edges and their included angle can be reproduced identically every time, without measuring every dimension - the basis of interchangeable parts.
Proof-building. SAS is a workhorse for proving later results: the base angles of isosceles triangles, properties of parallelograms, and many circle theorems all lean on a SAS step.
Surveying and navigation. When one distance is hard to reach, measuring two accessible sides and the angle between them fixes the whole triangle, letting the third distance be computed rather than walked.
The destination: once SAS is secure, you can prove triangles congruent in real diagrams and then transfer every matching part through CPCT - turning one established fact into six.
Common Mistakes With the SAS Criterion
Mistake 1: Using a non-included angle
Where it slips in: the problem gives two sides and an angle, and the angle is grabbed without checking its position.
Don't do this: apply SAS when the equal angle is $∠B$ but the equal sides are $AB$ and $AC$.
The correct way: confirm the equal angle is the one between the two equal sides. If it is not, you have SSA, which does not prove congruence.
Mistake 2: Mismatching corresponding parts
Where it slips in: naming the congruence with vertices in the wrong order.
Don't do this: write $\triangle ABC \cong \triangle EDF$ when $A$ corresponds to $D$, not $E$.
The correct way: list vertices so that corresponding parts line up: $A \leftrightarrow D$, $B \leftrightarrow E$, $C \leftrightarrow F$. A frequent slip is to write the second triangle's letters in alphabetical order out of habit rather than by correspondence.
Mistake 3: Treating SSA as if it were SAS
Where it slips in: two sides and a non-included angle look "close enough" to SAS.
Don't do this: claim congruence from SSA data.
The correct way: remember SSA can produce two different triangles from the same measurements - the "ambiguous case."
Practice Problems
Work these, then check below.
In $\triangle ABC$ and $\triangle XYZ$, $AB = XY$, $∠A = ∠X$, $AC = XZ$. Which rule proves congruence?
Two sides are 8 cm and 6 cm. In one triangle the included angle is $55°$; in another the angle between one of those sides and the third side is $55°$. Can SAS be applied?
In $\triangle PQR$, $PQ = PR$ and $PS$ bisects $∠P$ meeting $QR$ at $S$. Name the rule proving $\triangle PQS \cong \triangle PRS$.
True or false: SAS requires the third side of each triangle to be measured before congruence is confirmed.
Answer to Question 1: SAS - two sides and the included angle.
Answer to Question 2: No. In the second triangle the given angle is not included between the two given sides, so it is SSA, not SAS.
Answer to Question 3: SAS ($PQ = PR$, included $∠QPS = ∠RPS$, common side $PS$).
Answer to Question 4: False. SAS proves congruence from two sides and the included angle; the third side matches automatically.
Conclusion
The SAS criterion in triangles proves congruence from two sides and the included angle between them.
It can be justified by rigid transformations that map one triangle exactly onto another.
SAS works because the included angle removes all ambiguity; SSA fails because it does not.
Once SAS establishes congruence, all remaining sides and angles match by CPCT.
To master triangle congruence with a teacher, explore Bhanzu's geometry tutor or a high school math tutor, or browse math classes online.
A Practical Next Step
Now work through the four practice problems, checking that each equal angle you use is genuinely the included one. Want a live trainer to walk through a full SAS proof with you? Book a free demo class.
Read More
Was this article helpful?
Your feedback helps us write better content