AAS, HL, and Invalid Tests

Watch the video explainer

Learning goals

Derive AAS congruence from the triangle angle sum and ASA
Justify HL congruence using the isosceles base-angle theorem
Distinguish sufficient congruence information from AAA and unrestricted SSA data
Construct two noncongruent triangles satisfying one fixed set of SSA measurements

Two angles and a side not between them

The position of a known side need not stop a two-angle comparison. Suppose triangles $ABC$ and $DEF$ have the following corresponding measurements:

\begin{aligned} m\angle A&=43^\circ\\ m\angle D&=43^\circ\\ m\angle B&=66^\circ\\ m\angle E&=66^\circ\\ BC&=9\\ EF&=9.\end{aligned}

The known side is not between the two given angles.

The third angles match because both are what remains after subtracting the same two measures:

180^\circ-43^\circ-66^\circ=71^\circ.

Now use angles at $B,C$ and $E,F$ . The side of length $9$ lies between those pairs, so angle-side-angle (ASA) proves the triangles congruent. We changed which angles we used, not which measurements were given.

AAS follows from ASA#

Suppose two angles of one triangle match two corresponding angles of another, and one pair of corresponding nonincluded sides has equal lengths.

The third-angle theorem makes the remaining angles congruent. Thus all three angle pairs are available.

Take the two angles at the endpoints of the given equal sides. Those are corresponding angles, with the given equal sides between them. ASA proves the triangles congruent.

This is angle-angle-side (AAS): two corresponding angles and a corresponding side not between them establish congruence. As in the earlier criteria, the correspondence must stay consistent. We use nondegenerate triangles throughout: three noncollinear vertices and positive side lengths.

Worked example 1 Identify the side after matching the angles

Problem. In $\triangle JKL$ and $\triangle RST$ ,

\begin{aligned} \angle J&\cong\angle R \\ \angle L&\cong\angle T \\ JK&=15\text{ cm} \\ RS&=15\text{ cm}.\end{aligned}

Decide whether congruence follows and write the ordered statement.

The side marks and angle marks all use one vertex correspondence; the third vertices are unmarked.

Solution. The angles fix the first two vertex partners, and the remaining vertices must match:

\begin{aligned} J&\leftrightarrow R. \\ L&\leftrightarrow T. \\ K&\leftrightarrow S.\end{aligned}

The known side $JK$ corresponds to $RS$ under that same matching.

For the angles at $J,L$ , the included side would be $JL$ . The known side $JK$ is nonincluded, so AAS gives

\triangle JKL\cong\triangle RST.

Check your understanding

Two triangles have matching angles of 52 and 73 degrees and matching sides that are not between those angles. What lets you replace AAS with ASA?

A Assume a right angle B The third angles both measure 55 degrees C Every triangle with these angles has the same size D The two known angles must be equal to each other

A right angle changes the two-side case

In a right triangle, the side opposite the right angle is the hypotenuse, and the two sides meeting at the right angle are its legs. A right angle can make one hypotenuse and one leg enough to settle a congruence comparison.

The right angles are required information. They may be stated, marked, or established from perpendicular lines. A corner that merely looks square supplies no given.

For example, take two right triangles, each with a hypotenuse of length $17$ and a corresponding leg of length $8$ . Move one so the legs of length $8$ coincide, with both right-angle vertices at $A$ , placing their third vertices on opposite sides of that leg. In the diagram, those vertices are $C$ and $F$ , with the common leg $AB$ .

Align the equal legs. The right-angle givens place C, A, and F on one straight line.

The hypotenuses $BC$ and $BF$ , both $17$ , make the large triangle $BCF$ isosceles. Its base angles at $C,F$ match. Each small triangle also has a right angle at $A$ , so their remaining angles at $B$ match by the third-angle theorem. The known lengths are

\begin{aligned} AB&=8\\ BC&=17\\ BF&=17.\end{aligned}

The equal angles at $B$ are included between these sides, so SAS applies.

This works without calculating the other leg. Neither $8$ nor $17$ is essential to the argument.

The hypotenuse-leg criterion (HL) says that two right triangles are congruent when their hypotenuses and one pair of corresponding legs have equal lengths.

HL from the isosceles base-angle theorem#

Let $\triangle ABC$ and $\triangle DEF$ have right angles at $A,D$ , with

\begin{aligned} AB&=DE \\ BC&=EF.\end{aligned}

Translate and rotate the second triangle so $D$ lands on $A$ and $E$ on $B$ . If necessary, reflect across $AB$ so its third vertex, now named $F'$ , is on the opposite side of $AB$ from $C$ . Rigid motions preserve all the given lengths and angles.

Both $AC$ and $AF'$ are perpendicular to $AB$ at $A$ . They lie on the same perpendicular line in opposite directions, so $C,A,F'$ are collinear with $A$ between the other points.

The equal hypotenuses give

BC=BF'.

The base-angle theorem in $\triangle BCF'$ gives equal angles at $C$ and $F'$ . These are the angles at the third vertices of the two right triangles.

Their right angles also match, so their remaining angles $\angle ABC$ and $\angle ABF'$ are congruent. Use

\begin{aligned} AB&=AB \\ BC&=BF'.\end{aligned}

Together with those included angles, SAS gives

\begin{aligned} \triangle ABC&\cong\triangle ABF'.\end{aligned}

Reversing the alignment gives

\triangle ABC\cong\triangle DEF.

The angle marks record the base-angle and third-angle conclusions established in this proof. The other legs have no assumed equality.

Check your understanding

In this lesson's HL overlay, the right triangles share AB and have equal hypotenuses BC and BF. Which reason makes the angles at C and F equal before the angles at B are known to match?

A The third-angle theorem, using the right angles at A B The sides AC and AF have equal lengths C The base-angle theorem in triangle BCF, whose sides BC and BF are equal D The definition of a circle makes every angle formed by radii equal

Worked example 2 Check all three HL conditions

Problem. Triangles $PQR$ and $XYZ$ have right angles at $Q,Y$ . Given

\begin{aligned} PR&=XZ \\ PQ&=XY \\ QR&=22\text{ cm}.\end{aligned}

Decide whether congruence follows and find $YZ$ .

Only the supplied hypotenuse pair, leg pair, right angles, and the known 22 cm leg are marked.

Solution. The sides opposite the right angles are $PR,XZ$ , so the first equality matches the hypotenuses. The second equality matches the legs $PQ,XY$ . Both right angles are given.

HL gives the ordered congruence

\begin{aligned} \triangle PQR&\cong\triangle XYZ.\end{aligned}

The right-angle vertices match:

\begin{aligned} Q&\leftrightarrow Y.\end{aligned}

The other legs correspond, so

\begin{aligned} YZ&=QR \\ QR&=22\text{ cm} \\ YZ&=22\text{ cm} .\end{aligned}

Check your understanding

Two triangles have two pairs of equal corresponding sides. Before trying to use HL, which condition must be known about both triangles?

A The remaining side pair has unequal lengths B The triangles are drawn upright C Both triangles have a right angle D The two longest sides are perpendicular to each other

Three matching angles do not fix the size

Equal angles leave room for different side lengths. Construct an equilateral triangle with side length $7$ , then another with side length $18$ . Every angle in both triangles measures $60^\circ$ , but a rigid motion cannot turn a side of length $7$ into one of length $18$ .

Thus angle-angle-angle (AAA) is not a congruence criterion. It can hold for triangles that happen to be congruent, but it does not guarantee congruence.

The angles match in all three pairs. The stated side lengths prevent congruence.

Check your understanding

Two triangles have three matching angle pairs. Which conclusion is justified about congruence?

A They must be congruent B They cannot be congruent C Both triangles must have a right angle D They may or may not be congruent; the angles alone do not decide

The nonincluded angle can leave two triangles

Side-side-angle (SSA) gives two side lengths and an angle that is not between those sides. Unlike SAS, it is not sufficient in general.

Here is one fixed construction. Draw a ray from $A$ . From $A$ , draw a second ray making $30^\circ$ with it and mark $C$ at distance $10$ from $A$ . Draw a circle centered at $C$ with radius $7$ . It crosses the first ray at two points, $B_1$ and $B_2$ .

One circle supplies two choices for B. Both satisfy the same given side-side-angle data.

Both triangles have the same recorded lengths:

\begin{aligned} AC&=10 \\ CB_1&=7 \\ CB_2&=7.\end{aligned}

The two radius lengths are equal because they come from one circle. The angle at $A$ is also shared. That angle lies between $AC$ and the unknown base, not between the two given sides. The two distinct points on the same ray have different distances from $A$ , so

AB_1\ne AB_2.

The triangles share two side lengths but differ in the third, so their three-side lists cannot match under any vertex order. They are noncongruent.

This is a counterexample from a particular compass construction, not a general test for how many intersections every set of measurements has. SSA data can give no triangle, one triangle, or two. HL is a valid restricted case: the angle opposite the hypotenuse is known to be right, and the proof above settles that case.

Two matching side pairs and a nonincluded matching angle do not supply that right-angle condition. Without it, the general SSA ambiguity can remain.

Worked example 3 SAS or SSA depends on where the angle is

Problem. Compare each pair of nondegenerate triangles with the stated correspondence

\begin{aligned} A&\leftrightarrow D \\ B&\leftrightarrow E \\ C&\leftrightarrow F.\end{aligned}

In both cases

\begin{aligned} AB&=DE \\ AC&=DF.\end{aligned}

Case A. Also

\angle BAC\cong\angle EDF.

Does congruence follow?

Case B. Instead, only $\angle ABC\cong\angle DEF$ is known, with no right angles specified. Does this information guarantee congruence?

Both panels show triangle ABC. Top: Case A, with one arc at A. Bottom: Case B, with two arcs at B. The two known sides still meet at A.

Solution. Case A. The equal angles are between the two given side pairs. SAS gives

\triangle ABC\cong\triangle DEF.

Case B. The angles are at $B,E$ , while the two given sides meet at $A,D$ . This is unrestricted SSA. The circle construction gives a counterexample to its being a general criterion, so these data do not guarantee congruence.

Case B does not say that this particular pair must be noncongruent. It says the supplied information does not decide.

Check your understanding

In the two-intersection construction, why do both choices of B preserve the given length CB?

A Both points lie on the circle centered at C with the same radius B All sides on a ray have the same length C The triangles are congruent by SSA D Their unknown base lengths must be equal

AAS, HL, and Invalid Tests

Learning goals

Two angles and a side not between them

A right angle changes the two-side case

Three matching angles do not fix the size

The nonincluded angle can leave two triangles

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice

Your progress