SAS and ASA Congruence

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Learning goals

Identify the included angle between two sides and the included side between two angles
Justify SAS and ASA congruence by aligning triangles with rigid motions
Choose sufficient matching measurements and write the corresponding triangle congruence statement
Construct triangles from SAS or ASA data and explain the construction conditions

Which part lies between the others?

In the previous lesson, SSS used three matching side lengths to establish congruence. Congruence still means that a sequence of rigid motions maps the whole first triangle onto the second.

In triangle $ABC$ , sides $AB$ and $AC$ meet at $A$ . The angle they enclose, $\angle BAC$ , is their included angle. The angle at $B$ touches $AB$ , but it is not between $AB$ and $AC$ .

For angles at $A$ and $B$ , side $AB$ joins their vertices. It is their included side. Side $AC$ touches only one of those two vertices.

The two marked sides meet at the marked corner. This picture illustrates only the SAS pattern.

The marked side joins the two marked angle vertices. This separate picture illustrates the ASA pattern.

Check your understanding

Which angle is included between sides $LM$ and $MN$ ?

\angle MLN

\angle LMN

\angle LNM

D Any angle in the triangle

Two sides and the angle they enclose

Fix a side $AB$ of length $8$ cm. At $A$ , open a ray at $55^\circ$ to $AB$ , and place $C$ exactly $11$ cm along that ray. Once the side of line $AB$ is chosen, there is only one such point $C$ : moving it along the ray changes its distance from $A$ .

Now build $DEF$ with these measurements:

\begin{aligned} DE&=8\text{ cm} \\ DF&=11\text{ cm} \\ \angle EDF&=55^\circ \end{aligned}

Slide and turn the first triangle until $A$ lands on $D$ and $B$ on $E$ . If the two third vertices lie on opposite sides of the aligned base, reflect across that base. The $55^\circ$ angles now put $C$ and $F$ on the same ray from $D$ , and the $11$ cm lengths put them at the same point. The third sides match because both now join the same endpoints.

The side-angle-side criterion (SAS) says that two triangles are congruent if two pairs of corresponding sides and the pair of included angles are congruent. The angle must be between the two known sides in each triangle.

Why SAS gives a rigid-motion match#

The given triangles satisfy:

\begin{aligned} AB&=DE \\ AC&=DF \\ \angle BAC&\cong\angle EDF \end{aligned}

Translate $A$ to $D$ and rotate about $D$ to align ray $AB$ with ray $DE$ . Length preservation places $B$ on $E$ .

Call the image of $C$ point $C'$ . If necessary, reflect across line $DE$ so that $C'$ and $F$ are in the same half-plane. This fixes $D$ and $E$ and preserves the angle at $D$ . Equal angles opening from the same ray into that half-plane give the same second ray, by the angle-measure foundation accepted in Proving Lines Parallel or Perpendicular, lesson 2.3. Thus $C'$ lies on ray $DF$ . Distance preservation gives $DC'=DF$ .

On a ray, a positive distance from its endpoint selects one point, so $C'=F$ .

These are rigid motions, using the preservation properties accepted in chapter 3. The argument used the $8$ cm, $11$ cm, and $55^\circ$ example’s steps, without depending on those particular measurements. For a triangle the lengths are positive and the included angle is strictly between $0^\circ$ and $180^\circ$ ; a flat configuration is not a triangle.

After the first motions, the bases coincide. Reflection in that base fixes its endpoints and brings the third vertex onto its partner.

Worked example 1 Check the corner before naming the criterion

Problem. In triangles $JKL$ and $STU$ ,

\begin{aligned} JK&=6\text{ cm} \\ ST&=6\text{ cm} \\ KL&=10\text{ cm} \\ TU&=10\text{ cm} \\ \angle JKL&\cong\angle STU \end{aligned}

Establish congruence and write the matching order.

The ticks and arcs record the given matching pairs. Unmarked parts supply no additional data.

Solution. The known sides meet at $K$ in the first triangle and $T$ in the second. Those are the vertices of the known angles, so the angles are included.

The $6$ cm sides match $J$ with $S$ , and the $10$ cm sides match $L$ with $U$ . Together with $K\leftrightarrow T$ , this gives the ordered match by SAS:

\triangle JKL\cong\triangle STU.

Check your understanding

The given side pairs are:

\begin{aligned} AB&=PQ \\ BC&=QR \end{aligned}

Which additional equality directly completes SAS for $ABC$ and $PQR$ in that order?

\angle A\cong\angle P

\angle C\cong\angle R

\angle B\cong\angle Q

AB=QR

Two angles and the side joining them

Draw $UV=9$ cm.

Above the segment, a ray from $U$ makes $43^\circ$ with $UV$ , and a ray from $V$ makes $71^\circ$ with $VU$ . Their intersection gives the third vertex $W$ . The base fixes where the two rays start, and the angles fix their directions. The two distinct lines carrying the rays can meet at only one point.

If another triangle has that same base length and those same endpoint angles, align its base with $UV$ , putting its $43^\circ$ vertex on $U$ and its $71^\circ$ vertex on $V$ . Reflect in the base if needed so both third vertices are above it. Each corresponding ray then lies on the same ray as its partner. Their intersection is the same point, so the whole triangles coincide.

The angle-side-angle criterion (ASA) says that two triangles are congruent if two pairs of corresponding angles and the pair of included sides are congruent. Here $UV$ is included because it joins the vertices of the $43^\circ$ and $71^\circ$ angles.

Why ASA gives a rigid-motion match#

Suppose

\begin{aligned} \angle A&\cong\angle D \\ AB&=DE \\ \angle B&\cong\angle E \end{aligned}

Translate and rotate triangle $ABC$ to put $A$ on $D$ and $B$ on $E$ . If necessary, reflect across $DE$ to put its third vertex $C'$ in the same half-plane as $F$ .

Angle preservation puts ray $DC'$ on ray $DF$ and ray $EC'$ on ray $EF$ . The lines containing these rays are distinct, since the given figures are triangles. If they had two different intersection points, two distinct lines would pass through those same two points, contrary to the unique line through two distinct points. Thus $C'$ is $F$ .

As with the $9$ cm base and the $43^\circ$ and $71^\circ$ rays, the fixed base and its two endpoint directions leave one intersection. The reasoning does not depend on those numbers.

For ASA, aligning the base and choosing the same half-plane aligns both endpoint rays. The blue rays and red dashed image rays have the same single intersection.

Check your understanding

After aligning a base and putting the third vertices in the same half-plane, what fixes the third vertex in each justification?

A SAS uses the same ray and the same distance; ASA uses the unique intersection of two distinct ray lines B SAS uses two angles alone; ASA uses two side lengths alone C Both arguments work because the drawings have the same orientation D Both arguments require measuring the third side first

Worked example 2 Match the side to its two endpoint angles

Problem. In triangles $MNP$ and $XYZ$ ,

\begin{aligned} \angle M&\cong\angle Y \\ \angle P&\cong\angle X \\ MP&=12\text{ cm} \\ YX&=12\text{ cm} \end{aligned}

Establish congruence and write the first triangle in the order $MNP$ .

The two angle patterns identify the endpoints of the given side before the remaining vertices are matched.

Solution. Side $MP$ lies between the known angles at $M$ and $P$ ; side $YX$ lies between the matching angles at $Y$ and $X$ . ASA applies.

The full correspondence is:

\begin{aligned} M&\leftrightarrow Y \\ P&\leftrightarrow X \\ N&\leftrightarrow Z \end{aligned}

ASA gives the ordered statement:

\triangle MNP\cong\triangle YZX.

Check your understanding

In triangles $HJK$ and $RTS$ , angles at $H$ and $J$ match angles at $R$ and $T$ , respectively. Which side equality completes ASA for these angle pairs?

HJ=RT

HK=RS

JK=TS

D Any two side lengths are enough

Build the triangle from its data

A congruence criterion also explains why a construction fixes a triangle’s shape and size. Copying a length with a compass and copying a supplied angle are the procedures from chapter 1. The angle-copy procedure’s justification comes later in this chapter; here we use the copied angle as a specified input, just as we use a given angle in a proof.

Worked example 3 Construct from SAS data

Problem. You are given segments of lengths $5$ cm and $7$ cm and an angle template measuring $64^\circ$ . Construct a triangle with those two sides enclosing that angle. Explain why any successful construction has the same shape and size.

Solution. Draw $AB$ equal to the $7$ cm segment.

At $A$ , copy the supplied angle, using ray $AB$ as its first side. Choose either side of line $AB$ for the new ray.

Set the compass to the $5$ cm segment. Mark $C$ on the new ray at that distance from $A$ , then join $B$ to $C$ .

The copied angle fixes a ray; the compass setting locates C on that ray.

This gives

\begin{aligned} AB&=7\text{ cm} \\ AC&=5\text{ cm} \\ \angle BAC&=64^\circ \end{aligned}

Any other successful construction has the same two sides and included angle, so SAS establishes congruence. Choosing the other side of the base makes a reflected copy.

For SAS construction, both lengths must be positive and the included angle must lie strictly between $0^\circ$ and $180^\circ$ . These conditions make two rays that do not lie along one line. Positive lengths on them give three noncollinear vertices.

Worked example 4 Construct from ASA data

Problem. You are given a $10$ cm segment and angle templates measuring $46^\circ$ and $78^\circ$ . Construct a triangle whose known side joins those angles. Explain why the construction works and find the third angle.

Solution. Draw $DE=10$ cm.

At $D$ , copy the $46^\circ$ angle with one side along ray $DE$ . At $E$ , copy the $78^\circ$ angle with one side along ray $ED$ . Draw both new rays into the same half-plane bounded by line $DE$ .

Their angles toward that half-plane total

46^\circ+78^\circ=124^\circ.

Because this is less than $180^\circ$ , the rays converge on that side, by the parallel-postulate angle form from chapter 2. Name their intersection $F$ . Any other construction with the same data is congruent by ASA.

Both copied rays open into the same half-plane. Their intersection is the third vertex.

The triangle angle sum gives

\begin{aligned} \angle F&=180^\circ-124^\circ \\ \angle F&=56^\circ \end{aligned}

In general, ASA construction needs a positive base length and two positive interior angles whose sum is less than $180^\circ$ . For angles individually below $180^\circ$ , a sum of exactly $180^\circ$ makes the rays parallel; a larger sum does not produce an intersection in the chosen half-plane.

Check your understanding

Which data can construct a nonflat triangle with the given side between the two angles?

A A 6 cm side, angles

80^\circ

and

100^\circ

B A 6 cm side, angles

75^\circ

and

110^\circ

C A 0 cm side, angles

40^\circ

and

60^\circ

D A 6 cm side, angles

58^\circ

and

69^\circ

Choose the information before the criterion

Start with the proposed correspondence. Mark only what is given or already justified, then look for a complete pattern:

Criterion	Matching information	Placement check
SSS	Three side pairs	Each pair follows the same vertex correspondence
SAS	Two side pairs and an angle pair	The angles are between the known sides
ASA	Two angle pairs and a side pair	The sides join the known angle vertices

An angle drawn close to a marked side is not necessarily the included angle. Trace both sides from their shared vertex. Likewise, trace the segment joining two known angle vertices before claiming ASA. A missing mark is missing information, even when the sketch looks convincing.

SAS and ASA Congruence

Learning goals

Which part lies between the others?

Two sides and the angle they enclose

Two angles and the side joining them

Build the triangle from its data

Choose the information before the criterion

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice

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