Math Core

SSS and SAS lean mostly on sides. But in many figures, especially ones with parallel lines, angle bisectors or vertical angles, the easiest facts to find are angles. This lesson adds two shortcuts built mostly from angles, ASA and AAS, and finishes with a summary of which combinations work and which don't.

Angle-Side-Angle (ASA)

Just as an included angle sits between two sides, an included side sits between two angles: it's the side whose endpoints are the two angles' vertices. In $\triangle ABC$ , the included side of $\angle A$ and $\angle B$ is $\overline{AB}$ .

Imagine drawing a segment $\overline{AB}$ of fixed length, then drawing a ray from $A$ at a fixed angle and a ray from $B$ at a fixed angle. Two rays can cross in at most one point, so the third vertex is forced. Nothing about the triangle is left to choose.

ASA congruence

If two angles and the included side of one triangle are congruent to two angles and the included side of another triangle, then the triangles are congruent.

Two angles and the included side: ASA.

In the figure, $\angle A \cong \angle D$ , $\overline{AB} \cong \overline{DE}$ and $\angle B \cong \angle E$ . The marked side connects the two marked angles, so $\triangle ABC \cong \triangle DEF$ by ASA.

Angle-Angle-Side (AAS)

What if the congruent side is not between the two angles? It still works, and you can prove it with a tool from the first lesson of this unit.

Suppose $\angle A \cong \angle D$ , $\angle B \cong \angle E$ and $\overline{BC} \cong \overline{EF}$ . By the third angles theorem, $\angle C \cong \angle F$ . Now look at $\angle B$ , $\overline{BC}$ and $\angle C$ : that's two angles and the side between them. So the triangles are congruent by ASA. Because it can be proved from ASA, this shortcut is a theorem.

AAS congruence theorem

If two angles and a non-included side of one triangle are congruent to two angles and the corresponding non-included side of another triangle, then the triangles are congruent.

Two angles and a non-included side: AAS.

Common mistake

With AAS, the sides must correspond. The side has to be opposite the same angle in both triangles. If in one triangle the side is opposite the $40^\circ$ angle and in the other it's opposite the $85^\circ$ angle, the triangles are not congruent, even though you see "two angles and a side" in each.

What doesn't work: AAA and SSA

You've now seen four shortcuts. Two other combinations of three parts look similar but fail.

AAA. Three pairs of congruent angles fix a triangle's shape but not its size. A small equilateral triangle and a huge one both have three $60^\circ$ angles. (You'll study triangles like this in the Similarity unit.)
SSA. As you saw in the SSS and SAS lesson, two sides and a non-included angle can produce two different triangles.

Works	Doesn't work
SSS, SAS, ASA, AAS	AAA, SSA

Notice that every valid shortcut includes at least one side, and SSA is the only arrangement of two sides and one angle that fails.

Worked example: Matching the side to the right angles

$\triangle PQR$ has $m\angle P = 40^\circ$ , $m\angle Q = 85^\circ$ and $QR = 6$ . $\triangle XYZ$ has $m\angle X = 40^\circ$ , $m\angle Z = 55^\circ$ and $YZ = 6$ . Are the triangles congruent?

First find the missing angles. In $\triangle PQR$ , $m\angle R = 180^\circ - 40^\circ - 85^\circ = 55^\circ$ . In $\triangle XYZ$ , $m\angle Y = 180^\circ - 40^\circ - 55^\circ = 85^\circ$ .

So the angles match as $P \leftrightarrow X$ ( $40^\circ$ ), $Q \leftrightarrow Y$ ( $85^\circ$ ), $R \leftrightarrow Z$ ( $55^\circ$ ).

Now check the sides. $\overline{QR}$ is opposite $\angle P$ , and $\overline{YZ}$ is opposite $\angle X$ . Those angles correspond, so the sides correspond too, and both have length $6$ . The triangles are congruent by AAS: $\triangle PQR \cong \triangle XYZ$ . (Since you found all the angles, you could also use ASA with $\angle Q$ , $\overline{QR}$ , $\angle R$ .)

Proofs with ASA and AAS

Parallel lines are a rich source of congruent angles. When a transversal crosses two parallel lines, alternate interior angles are congruent.

Worked example: A proof using ASA

Given: $\overline{AB} \parallel \overline{CD}$ , and $E$ is the midpoint of $\overline{AD}$ .

Prove: $\triangle ABE \cong \triangle DCE$

AB is parallel to CD. Segments AD and BC cross at E, the midpoint of AD.

Statement	Reason
1. $\overline{AB} \parallel \overline{CD}$	Given
2. $\angle BAE \cong \angle CDE$	Alternate interior angles theorem
3. $E$ is the midpoint of $\overline{AD}$	Given
4. $\overline{AE} \cong \overline{DE}$	Definition of midpoint
5. $\angle AEB \cong \angle DEC$	Vertical angles are congruent
6. $\triangle ABE \cong \triangle DCE$	ASA

Check the order: the angle at $A$ , then side $\overline{AE}$ , then the angle at $E$ . The side is between the two angles, so ASA is the right name.

Worked example: A proof using AAS

Given: $\angle B \cong \angle D$ , and $\overline{AC}$ bisects $\angle BAD$ .

Prove: $\triangle ABC \cong \triangle ADC$

Diagonal AC bisects angle BAD, and angle B is congruent to angle D.

Statement	Reason
1. $\angle B \cong \angle D$	Given
2. $\overline{AC}$ bisects $\angle BAD$	Given
3. $\angle BAC \cong \angle DAC$	Definition of angle bisector
4. $\overline{AC} \cong \overline{AC}$	Reflexive property of congruence
5. $\triangle ABC \cong \triangle ADC$	AAS

In $\triangle ABC$ , the angles are at $B$ and $A$ , and the side $\overline{AC}$ is not between them (it runs from $A$ to $C$ ). So this is AAS, not ASA.

Tip

To decide between ASA and AAS, find the two vertices of the marked angles. If the marked side connects those two vertices, it's ASA. If not, it's AAS.

Practice

Practice 1

In $\triangle ABC$ and $\triangle DEF$ , $\angle A \cong \angle D$ , $\angle B \cong \angle E$ and $\overline{AB} \cong \overline{DE}$ . Which shortcut proves the triangles congruent?

Practice 2

In $\triangle ABC$ and $\triangle DEF$ , $\angle A \cong \angle D$ , $\angle B \cong \angle E$ and $\overline{BC} \cong \overline{EF}$ . Which shortcut proves the triangles congruent?

Practice 3

In $\triangle ABC$ and $\triangle DEF$ , $\angle A \cong \angle D$ , $\angle B \cong \angle E$ and $\angle C \cong \angle F$ . Which shortcut proves the triangles congruent?

Practice 4

Which of these is not a valid way to prove two triangles congruent?

Practice 5

In the proof that $\triangle ABE \cong \triangle DCE$ (where $\overline{AB} \parallel \overline{CD}$ ), what is the reason for $\angle BAE \cong \angle CDE$ ?

Practice 6

$\triangle ABC \cong \triangle DEF$ by ASA. $m\angle A = 52^\circ$ , $m\angle B = (3x + 5)^\circ$ and $m\angle E = 68^\circ$ . Find $m\angle F$ in degrees.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 7

$\angle B \cong \angle D$ and $\overline{AC}$ bisects $\angle BAD$ , so $\triangle ABC \cong \triangle ADC$ by AAS. If $m\angle BAC = (4x - 6)^\circ$ and $m\angle DAC = (2x + 18)^\circ$ , find $m\angle BAD$ in degrees.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 8

$\triangle KLM$ has $m\angle K = 35^\circ$ , $m\angle L = 100^\circ$ and $KM = 9$ . Which triangle must be congruent to $\triangle KLM$ ?