Triangle Congruence: SSS, SAS, ASA and AAS

Six parts, and why three are enough

A triangle has six parts: three sides and three angles. Two triangles are congruent when a sequence of rigid motions carries one onto the other.

Because rigid motions preserve distance and angle, congruent triangles have all six corresponding parts equal. The converse holds too — if all six match, a rigid motion exists.

$$\triangle ABC \cong \triangle DEF$$

The order of the letters is the correspondence: $A$ with $D$, $B$ with $E$, $C$ with $F$. So $AB = DE$, $\angle B = \angle E$, and so on for every part.

Checking all six parts is more work than needed. Three well-chosen parts force the other three, and the congruence criteria say which three.

The four criteria

Name	Given	“Included” means
SSS	three pairs of sides	—
SAS	two sides and the angle between them	the angle is formed by the two sides
ASA	two angles and the side between them	the side joins the two angles’ vertices
AAS	two angles and a side not between them	—

The word included matters. In SAS the angle must be the one formed by the two given sides; in ASA the side must be the one joining the two given angles.

Why SAS works

The criteria are not separate facts to memorize. Each one follows from rigid motions by the same argument.

Given $AB = DE$, $\angle A = \angle D$ and $AC = DF$.

1. Translate $\triangle ABC$ so that $A$ lands on $D$.
2. Rotate about $D$ until ray $AB$ lies along ray $DE$. Since $AB = DE$, the point $B$ lands exactly on $E$.
3. Ray $AC$ now makes an angle equal to $\angle D$ with ray $DE$. It lies either along ray $DF$ or along its mirror image across line $DE$. If it is the mirror image, reflect over line $DE$ — which leaves $D$ and $E$ fixed.
4. Ray $AC$ now lies along ray $DF$, and $AC = DF$, so $C$ lands on $F$.

All three vertices have landed on their partners. A sequence of rigid motions carried $\triangle ABC$ onto $\triangle DEF$, so they are congruent.

Every step was forced. The given parts left no choice about where each vertex could go — and that is exactly what a congruence criterion is.

Why SSS and ASA work

The first two steps are the same: move $A$ to $D$ and lay $AB$ along $DE$, reflecting if needed so $C$ and $F$ sit on the same side of line $DE$.

ASA. From $D$, the ray at angle $\angle A$ runs along ray $DF$. From $E$, the ray at angle $\angle B$ runs along ray $EF$. Point $C$ is where those two rays meet, and so is $F$. Two lines meet at most once, so $C = F$.

SSS. Point $C$ is at distance $AC = DF$ from $D$, so it lies on the circle around $D$ with that radius. It is also at distance $BC = EF$ from $E$. Two circles meet in at most two points, one on each side of line $DE$. $C$ and $F$ are on the same side, so $C = F$.

AAS reduces to ASA. Two known angles fix the third, since the three total $180°$, and then the known side is between two known angles.

Why SSA and AAA fail

AAA fixes the shape and leaves the size free. A triangle and its enlargement share all three angles and are similar, not congruent.

SSA is subtler. Take $AB = 10$, $\angle A = 30°$, and $BC = 5.5$ opposite that angle.

Swing a circle of radius $5.5$ around $B$. It crosses the other ray from $A$ in two places, making two different triangles — one obtuse, one acute — that both match every given part.

The solid triangle $ABC_1$ is obtuse, with an angle of about $115°$ at $C_1$. The dashed triangle $ABC_2$ is acute. Both have $AB = 10$, $\angle A = 30°$ and a third side of $5.5$ from $B$.

The given parts did not force a single position for the third vertex. A criterion works only when the given parts leave exactly one possibility, and SSA can leave two.

Finishing a proof with CPCTC

Congruence is usually proved not for its own sake but to learn something else. Once $\triangle ABC \cong \triangle DEF$, every corresponding part is equal:

CPCTC — Corresponding Parts of Congruent Triangles are Congruent.

Given $AB = CB$ and $\angle ABD = \angle CBD$, where $D$ is a point on $AC$. Prove $AD = CD$.

Statement	Reason
$AB = CB$	given
$\angle ABD = \angle CBD$	given
$BD = BD$	reflexive property
$\triangle ABD \cong \triangle CBD$	SAS
$AD = CD$	CPCTC

The shared side $BD$ is the step people miss. A side common to both triangles is equal to itself, and the reflexive property is the reason that says so.

Worked examples

Common mistakes

Practice problems