The ASA and AAS Congruence Criteria

If you fix two angles and any one side of a triangle, the third angle is forced (by the triangle angle sum) and the remaining sides are forced by trigonometric ratios. So two angles and a side uniquely determine a triangle , giving us two more congruence criteria: ASA and AAS.

Definitions

ASA Criterion (Angle–Side–Angle). If two angles and the included side of one triangle equal those of another, the triangles are congruent.

AAS Criterion (Angle–Angle–Side). If two angles and a non-included side of one triangle equal those of another, the triangles are congruent.

Both are valid because once you know two angles, you know the third (their sum is $180^\circ$); and one matching side then forces the others.

Why ASA works

Suppose $\triangle ABC$ and $\triangle DEF$ have $\angle B = \angle E$, $BC = EF$, and $\angle C = \angle F$. Place $\triangle ABC$ on $\triangle DEF$ so that $B$ falls on $E$ and $\overrightarrow{BC}$ on $\overrightarrow{EF}$. Since $BC = EF$, point $C$ falls on $F$. The angle $\angle B = \angle E$ means $\overrightarrow{BA}$ lies along $\overrightarrow{ED}$. The angle $\angle C = \angle F$ means $\overrightarrow{CA}$ lies along $\overrightarrow{FD}$. These two rays meet at exactly one point , so $A$ falls on $D$. The triangles coincide.

Why AAS works

If we have two angles and a non-included side, the third angle is determined by $\angle A + \angle B + \angle C = 180^\circ$. So AAS reduces to ASA after computing the third angle. The criterion is therefore valid.

In practice you can use AAS directly without computing the third angle , just state which two angles match and which (non-included) side matches, and conclude congruence.

Writing an ASA proof

Three lines:

1. State the first angle equality.
2. State the included side equality.
3. State the second angle equality, then conclude by ASA.

Example. In $\triangle ABC$ and $\triangle DEF$: $\angle A = \angle D$, $AB = DE$, $\angle B = \angle E$. Prove congruence.

Proof.

Writing an AAS proof

Identical structure , just identify which side is the non-included one and state AAS.

Example. $\triangle ABC$ and $\triangle DEF$: $\angle A = \angle D = 50^\circ$, $\angle B = \angle E = 60^\circ$, $BC = EF$. Prove congruence.

$\angle A, \angle B$ are the two matching angles; $BC$ is the side opposite $A$ (hence not included between $\angle A$ and $\angle B$, which would be $AB$). So this is AAS, and the triangles are congruent.

A classical application: angle bisector and a perpendicular

Claim. If $\overrightarrow{OP}$ is the angle bisector of $\angle XOY$ and from a point $P$ on the bisector, perpendiculars $PA$ and $PB$ are drawn to the two arms, then $PA = PB$.

Proof sketch. Triangles $\triangle OAP$ and $\triangle OBP$ share the side $OP$, have right angles at $A$ and $B$, and have equal angles at $O$ (since $OP$ bisects $\angle XOY$). So by AAS, the triangles are congruent. By CPCT, $PA = PB$.

This is a one-page proof that captures a beautiful symmetry property of angle bisectors.

Worked examples

Example 1. In $\triangle ABC$ and $\triangle DEF$: $\angle B = \angle E$, $BC = EF$, $\angle C = \angle F$. Justify congruence.

ASA (two angles with the included side).

Example 2. In $\triangle ABC$ and $\triangle DEF$: $\angle A = \angle D$, $\angle B = \angle E$, $AC = DF$. Justify congruence.

$AC$ is the side opposite $B$ , not included between $A$ and $B$. So this is AAS.

Example 3. Two triangles have $\angle A = \angle P = 50^\circ$, $\angle B = \angle Q = 60^\circ$. Are they necessarily congruent?

No. AAA is not a congruence criterion , it gives similarity but not congruence (the triangles might be different sizes).

Example 4. $\triangle ABC$ has $\angle A = 50^\circ, \angle B = 60^\circ$. $\triangle PQR$ has $\angle P = 50^\circ, \angle Q = 60^\circ$, and $PQ = AB$. Are the triangles congruent?

$\angle A = \angle P$ (both $50^\circ$), $AB = PQ$, $\angle B = \angle Q$ (both $60^\circ$). The matching side $AB = PQ$ is included between $\angle A$ and $\angle B$. So ASA $\Rightarrow$ congruent.

Example 5. Lines $\ell$ and $m$ are parallel, cut by a transversal making alternate-interior angles at $P$ on $\ell$ and $Q$ on $m$. If $R$ is the midpoint of $PQ$, prove that any line through $R$ makes triangles on the two sides that are congruent by ASA.

The alternate-interior angles equal each other (so two angles match), and $PR = QR$ (midpoint). Hence ASA.

Try it yourself

1. State the ASA criterion.
2. State the AAS criterion.
3. Why does AAS work given AAA does not?
4. In $\triangle PQR$ and $\triangle STU$: $\angle P = \angle S$, $PQ = ST$, $\angle Q = \angle T$. Are the triangles congruent? Which criterion?
5. Two triangles share an angle and have two pairs of equal angles. Can we conclude they are congruent?
6. From a point on the bisector of an angle, perpendiculars are drawn to the two arms. Prove the perpendiculars are equal in length using AAS.
7. In $\triangle ABC$, $D$ is the foot of the perpendicular from $A$ to $BC$. If $\angle B = \angle C$, prove $\triangle ABD \cong \triangle ACD$.
8. State whether AAA is a congruence criterion.
9. Why is the "included side" important in ASA?
10. In a triangle, two angles are $40^\circ$ and $70^\circ$. State the third.

Pitfalls / Insight

• AAA gives only similar triangles, not congruent. A side must always be matched in some way.
• Identify "included" vs "non-included" precisely. ASA needs the side between the two matched angles.
• CPCT after the conclusion. Once you cite ASA or AAS, immediately have access to all six corresponding parts.

Insight. Three congruence criteria , SAS, ASA, AAS , handle the great majority of triangle-congruence proofs. Once you can spot which one a figure offers, you write the three-line proof and finish with a CPCT remark.