Triangle Congruence and SSS

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Learning goals

Match corresponding vertices, sides, and angles from an ordered triangle congruence statement
Decide whether three pairs of side lengths establish the stated triangle congruence
Explain how rigid motions justify the SSS criterion, including the reflected case
Construct a triangle from three lengths and explain its possible placements

Read the match in order

Two figures are congruent when a sequence of rigid motions carries one exactly onto the other. A slide, turn, or reflection changes its position but preserves lengths and angle measures, as established in Compositions and Congruence.

The ordered statement is

\begin{aligned} \triangle ABC&\cong\triangle RST. \end{aligned}

It specifies a match: $A$ goes to $R$ , $B$ to $S$ , and $C$ to $T$ . These are corresponding vertices, the vertices paired by the mapping. The order matters even when a drawing is rotated or reflected.

Follow the endpoints to find corresponding sides. For example, $BC$ pairs with $ST$ , not with $RT$ . Follow the middle letter to match an angle: $\angle BAC$ pairs with $\angle SRT$ .

Worked example 1 Read an ordered statement

Problem. The given ordered statement is

\begin{aligned} \triangle ABC&\cong\triangle RST. \end{aligned}

Name the side corresponding to $AC$ and the angle corresponding to $\angle BCA$ . The measurements are

\begin{aligned} ST&=9\text{ cm}. \\ m\angle R&=47^\circ. \end{aligned}

Find $BC$ and $m\angle A$ .

Read the given order across the turn. Only ST and angle R carry the stated measurements; the drawing is schematic.

Solution. The vertex pairs are

\begin{aligned} A&\leftrightarrow R \\ B&\leftrightarrow S \\ C&\leftrightarrow T. \end{aligned}

Thus $AC$ corresponds to $RT$ , and $\angle BCA$ corresponds to $\angle STR$ .

The length $BC$ equals its corresponding length $ST$ , so

\begin{aligned} BC&=9\text{ cm}. \end{aligned}

Angle $A$ corresponds to angle $R$ , so

\begin{aligned} m\angle A&=47^\circ. \end{aligned}

Check your understanding

If triangle KLM is congruent to triangle VUW in that order, which side corresponds to KM?

A VU B UW C VW D KM

For congruent triangles, all six corresponding parts match: three pairs of sides and three pairs of angles. The matching motions carry each side and angle onto its partner and preserve lengths and angle measures.

The side pairs are:

\begin{aligned} AB&=RS. \\ BC&=ST. \\ CA&=TR. \end{aligned}

The angle pairs are:

\begin{aligned} m\angle A&=m\angle R. \\ m\angle B&=m\angle S. \\ m\angle C&=m\angle T. \end{aligned}

Conversely, if all six pairs match in one consistent correspondence, the triangles are congruent. This follows from SSS below, which shows that matching just the three side pairs already suffices. Equal lengths use $=$ ; congruent segments or triangles use $\cong$ .

What three sides determine

Three matching side lengths are enough to force an exact triangle match. Imagine one triangle with side lengths $4$ , $6$ , and $8$ cm and another with the same three lengths. Slide and turn the first so the two $8$ cm sides coincide, matching their endpoints. Its third vertex must be $4$ cm from one endpoint and $6$ cm from the other.

Those distance requirements place the vertex on two circles. There are two possible positions, one on each side of the common base. They are mirror images; reflecting across the base exchanges them. Either position gives an exact match after a rigid motion. The next section justifies why there are no extra positions hidden in this description.

The Side-Side-Side criterion, abbreviated SSS, says: if all three side lengths of one triangle equal their corresponding side lengths in another, the triangles are congruent. Here and throughout, a triangle has three distinct, noncollinear vertices. A straight chain of three segments is not a triangle.

Worked example 2 Check sides and the vertex order

Problem. Triangle $JKL$ has these side lengths in centimeters:

\begin{aligned} JK&=10. \\ KL&=14. \\ JL&=16. \end{aligned}

Triangle $UVW$ has these side lengths in centimeters:

\begin{aligned} UV&=14. \\ VW&=16. \\ UW&=10. \end{aligned}

Establish a correct ordered congruence statement.

Every length label is a given. Use the two lengths meeting at a vertex to identify its partner.

Solution. Because all three side lengths differ, the two lengths meeting at each vertex identify exactly one partner. Vertex $J$ joins the $10$ and $16$ cm sides. In the second triangle, $W$ joins those lengths, so $J$ corresponds to $W$ .

Vertex $K$ joins the $10$ and $14$ cm sides, so it corresponds to $U$ . Vertex $L$ therefore corresponds to $V$ . Check each pair:

\begin{aligned} JK&=WU \\ JK&=10 \\ KL&=UV \\ KL&=14 \\ JL&=WV \\ JL&=16 \end{aligned}

The correct order, by SSS, is

\begin{aligned} \triangle JKL&\cong\triangle WUV. \end{aligned}

Worked example 3 Complete the side data

Problem. A triangle has side lengths $11$ , $13$ , and $17$ cm. A second actual triangle has corresponding sides $11$ , $13$ , and $2x+5$ cm. With these sides corresponding in the stated order, find $x$ so that SSS establishes congruence, and check the third side.

The third side pair is unfinished. The schematic drawings supply no value for x.

Solution. The first two pairs already match. The remaining lengths must be equal:

\begin{aligned} 2x+5&=17 \end{aligned}

Subtract $5$ and divide by $2$ , giving

\begin{aligned} x&=6. \end{aligned}

Substitute to check the third side:

\begin{aligned} 2(6)+5&=17\text{ cm}. \end{aligned}

Thus all three pairs match.

Check your understanding

Triangles ABC and DEF have these corresponding side lengths:

\begin{aligned} AB&=DE. \\ AB&=12. \\ BC&=EF. \\ BC&=15. \\ AC&=DF. \\ AC&=19. \end{aligned}

Which conclusion follows?

A Triangle ABC is congruent to triangle DEF by SSS B Only two side pairs match C The triangles need matching positions first D Their angle measures must be measured separately before congruence follows

Why the rigid motions finish the job

Matching a base is easy: translate one endpoint to its partner and rotate the base onto its partner. Equal base lengths make the other endpoints coincide too. The remaining issue is the third vertex. We can settle it using the Pythagorean theorem from Pre-Algebra, which says that the squares of a right triangle’s leg lengths add to the square of its hypotenuse. We use that previously learned fact here; its geometric proof will return later in this course.

Three matching sides guarantee triangle congruence#

First use lengths $8$ , $7$ , and $9$ . Place the base endpoints at $(0,0)$ and $(8,0)$ . Call a possible third vertex $(x,y)$ , with distance $7$ from the first endpoint and $9$ from the second. Dropping a perpendicular to the horizontal axis gives right triangles, so the prior Pythagorean fact gives

\begin{aligned} x^2+y^2&=49 \\ (x-8)^2+y^2&=81 \end{aligned}

The squares also work if the perpendicular foot lies beyond an endpoint: a negative coordinate has the same square as its positive length. If the foot is an endpoint, that horizontal distance is zero and the vertical distance alone gives the corresponding side length, so its squared equation still holds. Subtract the first equation from the second:

\begin{aligned} 64-16x&=32 \end{aligned}

Thus

\begin{aligned} x&=2. \end{aligned}

Substitution in the first equation gives

\begin{aligned} y^2&=45. \end{aligned}

The only possible vertices are

\begin{aligned} \text{One vertex: }&(2,\sqrt{45}). \\ \text{Other vertex: }&(2,-\sqrt{45}). \end{aligned}

Reflection across the horizontal base exchanges these points and fixes both base endpoints.

For any two actual triangles with matching side lengths, place their common base at $(0,0)$ and $(d,0)$ , where

\begin{aligned} d&>0. \end{aligned}

Let the distances from a third vertex $(x,y)$ to the endpoints be $r$ and $s$ .

The perpendicular foot gives the horizontal and vertical legs used in the distance equations. This drawing shows a foot between the base endpoints.

The same reasoning gives

\begin{aligned} x^2+y^2&=r^2 \\ (x-d)^2+y^2&=s^2 \end{aligned}

Subtracting fixes the horizontal coordinate:

\begin{aligned} d^2-2dx&=s^2-r^2 \\ x&=\frac{d^2+r^2-s^2}{2d} \end{aligned}

The first equation then fixes $y^2$ . Because these are noncollinear triangles,

\begin{aligned} y^2&>0. \end{aligned}

The two possible heights have opposite signs and equal size. A reflection across the base changes one to the other. No part of this reasoning depended on choosing $8$ , $7$ , or $9$ .

After the base-matching translation and rotation, either the third vertices already coincide or that reflection makes them coincide. A rigid motion carries each segment onto the segment joining its image endpoints, so all three sides coincide. The whole triangles match.

After the base is matched, only the side of the base may differ. A reflection fixes both base endpoints and exchanges the third vertices.

This argument uses the accepted rigid-motion preservation properties, coordinates, and the Pythagorean theorem from Pre-Algebra. The circle picture locates candidates; the argument proves that they differ at most by reflection.

Check your understanding

After corresponding bases coincide, two third vertices are on opposite sides of the base with the same horizontal position and equal heights in size. Which motion completes the match while leaving the base fixed?

A A stretch away from the base B A translation parallel to the base C A quarter-turn about one endpoint D A reflection across the base

Construct a triangle from its sides

A compass can locate a point by distance without measuring an angle. To construct $\triangle PQR$ , use these given lengths:

\begin{aligned} PQ&=9\text{ cm}. \\ PR&=6\text{ cm}. \\ QR&=7\text{ cm}. \end{aligned}

Draw $PQ$ of length $9$ cm.
Draw an arc centered at $P$ with radius $6$ cm.
Draw an arc centered at $Q$ with radius $7$ cm so it crosses the first arc.
Choose an intersection as $R$ and join $PR$ and $QR$ .

Every point on the first arc is $6$ cm from $P$ , and every point on the second is $7$ cm from $Q$ . Their intersection satisfies both requirements. Choosing the other intersection gives a reflected, congruent triangle by SSS, not a new triangle shape.

The upper and lower arc intersections give reflected copies of the same triangle. The arcs keep the given radii.

Not every three positive lengths produce a triangle. The arcs might miss or merely touch. For example, lengths $2$ , $3$ , and $5$ cm place the third vertex on the line through the base when the arcs touch, so there is no noncollinear triangle. We will develop a complete test for possible side lengths in the next chapter.

Check your understanding

To construct triangle DEF, the given lengths are

\begin{aligned} DE&=10\text{ cm}. \\ DF&=8\text{ cm}. \\ EF&=9\text{ cm}. \end{aligned}

Which arcs locate F after DE is drawn?

A Two arcs centered at D with radii 8 and 9 B An arc centered at D of radius 8 and one centered at E of radius 9 C An arc centered at D of radius 9 and one centered at E of radius 8 D Two arcs of radius 10

Why a four-sided frame can move

SSS is a statement about triangles. Four fixed side lengths do not generally determine a four-sided figure. A square made from four equal hinged strips can lean into a slanted frame without changing any strip. Its angles change, so the new frame is not congruent to the square.

Each frame uses four strips of the same length. Matching four side lengths leaves room for different angles.

Adding a diagonal divides the frame into triangles. Fixing that extra length can stop this leaning, because the pieces must now satisfy three side lengths each. This explains why a diagonal brace can make a frame rigid.

Triangle Congruence and SSS

Learning goals

Read the match in order

What three sides determine

Why the rigid motions finish the job

Construct a triangle from its sides

Why a four-sided frame can move

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice

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