Math Core

To show two triangles are congruent from the definition, you'd have to check six pairs of parts. That's a lot of work, and luckily it's more than you need. In this lesson you'll learn the first two congruence shortcuts, SSS and SAS, which prove two triangles congruent from just three well-chosen pairs of parts.

Side-Side-Side (SSS)

Try this with three straws cut to lengths $3$ , $4$ and $5$ inches. However you connect their ends, you always get the same triangle. You can flip it or turn it, but you can't change its shape. This is why triangles are used in bridges, roof trusses and bike frames: once the three side lengths are fixed, the triangle is rigid. (Four straws joined in a loop, by contrast, can be squashed into many different shapes.)

SSS congruence

If three sides of one triangle are congruent to three sides of another triangle, then the triangles are congruent.

Three pairs of congruent sides: SSS.

In the figure, $\overline{AB} \cong \overline{DE}$ , $\overline{BC} \cong \overline{EF}$ and $\overline{CA} \cong \overline{FD}$ , so $\triangle ABC \cong \triangle DEF$ by SSS. You never had to measure an angle; the angles are forced to match.

Side-Angle-Side (SAS)

The second shortcut uses two sides and one angle, but the angle has to be in a particular place.

Definition

Included angle

The included angle of two sides of a triangle is the angle formed by those two sides. Its vertex is the point where the sides meet. In $\triangle ABC$ , the included angle of $\overline{AB}$ and $\overline{AC}$ is $\angle A$ .

Picture a door hinge. If you fix the lengths of two boards joined at a hinge and also fix the angle of the hinge, the distance between the free ends is locked in. That means the third side is determined, and so is the whole triangle.

SAS congruence

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

Two pairs of sides and the included angle: SAS.

Here $\overline{AB} \cong \overline{DE}$ , $\angle A \cong \angle D$ and $\overline{AC} \cong \overline{DF}$ . The angle sits between the two marked sides, so $\triangle ABC \cong \triangle DEF$ by SAS.

Common mistake

The angle must be included. Two sides and an angle that is not between them (sometimes called SSA) do not guarantee congruence. In the figure below, both triangles have side $\overline{AB}$ , angle $A$ , and a third side of the same length, yet one triangle is clearly bigger. The side opposite $A$ can swing into two different positions.

Triangles ABC₁ and ABC₂ share angle A and side AB, and BC₁ = BC₂. They are not congruent.

Finding hidden information

Proofs often use facts that aren't marked on the diagram but follow from it. Look for these:

A shared side. A side that belongs to both triangles is congruent to itself by the reflexive property of congruence.
Vertical angles. When two segments cross, the vertical angles are congruent.
Midpoints and bisectors. A midpoint splits a segment into two congruent pieces; an angle bisector splits an angle into two congruent angles.
Perpendicular lines. They form right angles, and all right angles are congruent.

Proofs with SSS and SAS

A congruence proof ends with a triangle congruence statement, and its reason is the name of the shortcut. Each piece of the shortcut needs its own line with a reason.

Worked example: A proof using SAS

Given: $M$ is the midpoint of $\overline{AE}$ and of $\overline{BD}$ .

Prove: $\triangle AMB \cong \triangle EMD$

Segments AE and BD cross at M, the midpoint of both.

Statement	Reason
1. $M$ is the midpoint of $\overline{AE}$ and of $\overline{BD}$	Given
2. $\overline{AM} \cong \overline{EM}$	Definition of midpoint
3. $\angle AMB \cong \angle EMD$	Vertical angles are congruent
4. $\overline{BM} \cong \overline{DM}$	Definition of midpoint
5. $\triangle AMB \cong \triangle EMD$	SAS

The angle in step 3 is formed by $\overline{AM}$ and $\overline{BM}$ in one triangle and by $\overline{EM}$ and $\overline{DM}$ in the other, so it really is the included angle.

Worked example: A proof using SSS and a shared side

Given: $\overline{AB} \cong \overline{CB}$ and $\overline{AD} \cong \overline{CD}$

Prove: $\triangle ABD \cong \triangle CBD$

Quadrilateral ABCD with diagonal BD.

Statement	Reason
1. $\overline{AB} \cong \overline{CB}$	Given
2. $\overline{AD} \cong \overline{CD}$	Given
3. $\overline{BD} \cong \overline{BD}$	Reflexive property of congruence
4. $\triangle ABD \cong \triangle CBD$	SSS

Diagonal $\overline{BD}$ is a side of both triangles, which supplies the third pair of sides.

SSS on the coordinate plane

The distance formula (or just counting, for horizontal and vertical sides) lets you compare side lengths directly.

Worked example: Checking SSS with coordinates

$\triangle PQR$ has vertices $P(0, 0)$ , $Q(3, 0)$ , $R(0, 4)$ . $\triangle STU$ has vertices $S(5, 1)$ , $T(5, 4)$ , $U(1, 1)$ . Are the triangles congruent?

Find the side lengths.

\begin{aligned} PQ &= 3, & PR &= 4, & QR &= \sqrt{(0 - 3)^2 + (4 - 0)^2} = \sqrt{25} = 5 \\ ST &= 3, & SU &= 4, & TU &= \sqrt{(1 - 5)^2 + (1 - 4)^2} = \sqrt{25} = 5 \end{aligned}

All three pairs match, so the triangles are congruent by SSS. Matching the sides gives the correspondence $P \leftrightarrow S$ , $Q \leftrightarrow T$ , $R \leftrightarrow U$ , so $\triangle PQR \cong \triangle STU$ .

Tip

Before naming a shortcut, list the three pairs you actually have, in order around the triangle. If you have side, angle, side with the angle in the middle, it's SAS. If the angle is at the end of the list, you don't have SAS.

Practice

Practice 1

Which shortcut proves the triangles congruent?

Triangles ABC and DEF. Matching marks show congruent parts.

Practice 2

In $\triangle ABC$ and $\triangle DEF$ , $\overline{AB} \cong \overline{DE}$ , $\overline{BC} \cong \overline{EF}$ and $\angle A \cong \angle D$ . Are the triangles necessarily congruent?

Practice 3

In $\triangle ABC$ and $\triangle DEF$ , you know $\overline{AB} \cong \overline{DE}$ and $\angle B \cong \angle E$ . What else do you need to prove $\triangle ABC \cong \triangle DEF$ by SAS?

Practice 4

In the proof that $\triangle AMB \cong \triangle EMD$ (where $M$ is the midpoint of $\overline{AE}$ and of $\overline{BD}$ ), what is the reason for the statement $\angle AMB \cong \angle EMD$ ?

Practice 5

$\overline{PQ} \cong \overline{RQ}$ , and $\overline{QS}$ bisects $\angle PQR$ , with $S$ on $\overline{PR}$ . Which shortcut proves $\triangle PQS \cong \triangle RQS$ ?

Practice 6

One triangle has sides of length $6$ , $10$ and $12$ . Another has sides of length $6$ , $10$ and $3x - 3$ . What value of $x$ makes the triangles congruent by SSS?

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

Practice 7

$\triangle ABC$ has vertices $A(0, 0)$ , $B(6, 0)$ and $C(0, 8)$ . $\triangle DEF$ has $DE = 6$ , $DF = 8$ and $m\angle D = 90^\circ$ . Explain why the triangles are congruent, then find $EF$ .

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

Practice 8

Quadrilateral $WXYZ$ has $\overline{WX} \cong \overline{YZ}$ and $\overline{XY} \cong \overline{ZW}$ . Diagonal $\overline{WY}$ is drawn. Which statement is proved by SSS?