Congruence of Triangles and the SAS Criterion

Two triangles are congruent when one can be superimposed exactly on the other , every side matches, every angle matches. In practice we do not always slide and rotate triangles physically; instead we look for enough matching parts to guarantee a perfect overlap. The first such guarantee is the SAS criterion: matching two sides and the angle between them.

Definitions

Two triangles $\triangle ABC$ and $\triangle DEF$ are congruent, written $\triangle ABC \cong \triangle DEF,$ if their six pairs of corresponding parts are equal: three pairs of sides $AB = DE, \quad BC = EF, \quad CA = FD,$ and three pairs of angles $\angle A = \angle D, \quad \angle B = \angle E, \quad \angle C = \angle F.$

The order of vertices matters. The notation $\triangle ABC \cong \triangle DEF$ specifies a correspondence: $A \leftrightarrow D$, $B \leftrightarrow E$, $C \leftrightarrow F$. Writing $\triangle ABC \cong \triangle EDF$ instead asserts a different correspondence and would usually be wrong.

A useful shorthand: CPCT , Corresponding Parts of Congruent Triangles. Once you prove two triangles congruent, you can invoke "CPCT" to claim that any pair of corresponding parts is equal, without re-proving it.

The SAS criterion

You do not need all six pairs of equalities to prove congruence , three carefully chosen ones suffice.

SAS Criterion (Side–Angle–Side). If two sides and the included angle of one triangle are respectively equal to two sides and the included angle of another triangle, then the two triangles are congruent.

In symbols: in $\triangle ABC$ and $\triangle DEF$, if $AB = DE$, $\angle A = \angle D$, and $AC = DF$, then $\triangle ABC \cong \triangle DEF$.

Key word: "included". The angle must lie between the two sides. SAS does not work if the angle is opposite one of the matching sides; that arrangement is called "SSA" (or "ASS") and does not in general guarantee congruence.

Why does SAS work?

Geometrically, two sides and the included angle determine the triangle uniquely: pin one vertex, draw two segments of specified lengths at a specified angle, and connect the endpoints. There is no choice; the third side is forced.

In Euclid's Elements, SAS is taken as an axiom (Proposition 4). We accept it here without further proof and use it as a building block.

Writing an SAS proof

A standard SAS proof has three lines:

1. State the two matching sides.
2. State the matching included angle.
3. Conclude by SAS.

Example. In $\triangle ABC$ and $\triangle DEF$: $AB = DE$, $\angle A = \angle D$, and $AC = DF$. Prove the triangles are congruent.

Proof.

After the conclusion, any corresponding part can be invoked via CPCT. For instance: $BC = EF$ by CPCT. Q.E.D.

Two classical applications

Application 1: The diagonals of a rhombus bisect its angles. This is proved by showing two triangles formed by a diagonal are congruent via SAS , the diagonal is a shared side, two adjacent sides are equal (it's a rhombus), and the angles between them are equal.

Application 2: An isosceles triangle has equal base angles. Reflect across the perpendicular from the apex to the base; the two half-triangles match by SAS, and the base angles correspond , hence they are equal. We give this proof carefully in lesson 4.

Worked examples

Example 1. In $\triangle ABC$ and $\triangle PQR$, $AB = PQ = 5$, $\angle A = \angle P = 60^\circ$, $AC = PR = 7$. Are the triangles congruent?

Yes, by SAS. (Two sides and the included angle.)

Example 2. In $\triangle ABC$ and $\triangle PQR$, $AB = PQ = 5$, $BC = QR = 6$, $\angle A = \angle P = 50^\circ$. Are these triangles necessarily congruent?

Be careful: the angle is at $A$, between sides $AB$ and $AC$ , not between $AB$ and $BC$. So $\angle A$ is opposite $BC$, not between $AB$ and $BC$. The data is "SSA", which does not guarantee congruence. We cannot conclude.

Example 3. Two segments $AB$ and $CD$ bisect each other at $O$. Prove $\triangle AOC \cong \triangle BOD$.

$AO = BO$ (bisection), $CO = DO$ (bisection), $\angle AOC = \angle BOD$ (vertically opposite). SAS $\Rightarrow$ congruent.

Example 4. In $\triangle ABC$, $AB = AC$ and $\overline{AD}$ is the median to $BC$. Prove $\triangle ABD \cong \triangle ACD$.

$AB = AC$ (given), $BD = CD$ (median bisects $BC$), and $AD = AD$ (common). This uses SSS , but we can also do SAS: $AB = AC$, $\angle ABD = \angle ACD$ (later, by isosceles theorem , we'll postpone this argument). For now, with the median and SSS: $\triangle ABD \cong \triangle ACD$.

Example 5. Two triangles $\triangle ABC$ and $\triangle DEF$ have $AB = DE$, $BC = EF$, $\angle B = \angle E$. Conclude.

Two sides and the included angle (at $B$ and $E$). SAS $\Rightarrow \triangle ABC \cong \triangle DEF$.

Try it yourself

1. State the SAS criterion in your own words.
2. Why must the angle in SAS be the included angle?
3. In $\triangle PQR$ and $\triangle STU$, $PQ = ST = 4$, $\angle Q = \angle T = 90^\circ$, $QR = TU = 5$. Are they congruent?
4. Why is "SSA" not a valid criterion?
5. Two line segments $AB$ and $CD$ intersect at $O$, and $O$ is the midpoint of each. Prove $\triangle AOC \cong \triangle BOD$.
6. State CPCT in your own words.
7. In $\triangle ABC \cong \triangle DEF$, list all six corresponding parts that are equal.
8. Sketch two triangles that satisfy SAS , choose your own values.
9. In a rhombus $ABCD$, prove $\triangle ABD \cong \triangle CBD$ using SAS.
10. In $\triangle ABC$, $D$ is on $BC$ such that $BD = DC$. If $\angle BAD = \angle CAD$, can we conclude $\triangle ABD \cong \triangle ACD$? Justify.

Pitfalls / Insight

• Order of vertices in the congruence statement. $\triangle ABC \cong \triangle DEF$ means $A \leftrightarrow D$, $B \leftrightarrow E$, $C \leftrightarrow F$. Writing the wrong order is a common error.
• "Included" angle is non-negotiable. SSA is not a criterion in general.
• CPCT is your gold mine. Once two triangles are congruent, you immediately have six equalities.

Insight. SAS is the first and most direct congruence test. Watch for two pairs of equal sides with a shared or matching angle in between, and the rest of the proof writes itself.