Geometry · G.CO.B.7

Congruence statements and correspondence

Two triangles are exactly the same size and shape. When you write that down, how does the reader know which corner of one goes with which corner of the other?

Unlocks

Proving parts congruent after the trianglesGeometry

Where you are 3 in Congruent triangles

Geometry11

Triangles18

Congruent triangles3

Congruence statements and correspondence

Why AAA and SSA fail Hypotenuse leg congruence

When you reach for it

Look at the two triangles in Figure 1. The one on the right is the one on the left, flipped over and turned. They are the same size and the same shape, so they are congruent. But saying “the triangles are congruent” leaves out the part you actually need: which side of one matches which side of the other, and which angle matches which angle.

Figure 1Six matching marks, one for each pair of parts. The statement \triangle ABC \cong \triangle DEF says the same thing in one line.

A triangle has six parts, three sides and three angles. To say two triangles are congruent is to say that their parts can be paired off, vertex to vertex, so that all six pairs match.

Definition. Two triangles are congruent when their vertices can be matched one to one so that each pair of matched sides is congruent and each pair of matched angles is congruent.

The matching of vertices is called a correspondence. In Figure 1 it is

A \leftrightarrow D \qquad B \leftrightarrow E \qquad C \leftrightarrow F

and the marks show the six pairs it produces: matching ticks for sides, as for congruent segments, and matching arcs for angles, as for congruent angles.

Writing out six pairs every time would be slow, so geometry packs the whole correspondence into the order of the letters:

\triangle ABC \cong \triangle DEF

This is a congruence statement. The first letter on the left goes with the first letter on the right, the second with the second, the third with the third. One line holds all six facts, and a reader can list them without seeing the figure.

You reach for this whenever two triangles are in play: to record what a figure shows, to read off a missing side or angle, and later to say exactly what a proof has established. The same matching can be pictured as rigid motions carrying each vertex onto its partner; this lesson works with the letters.

The procedure

Match the letters by position. In \triangle ABC \cong \triangle DEF the positions give the correspondence A \leftrightarrow D, B \leftrightarrow E, C \leftrightarrow F. That is the whole rule, and it is a convention: \cong between two triangle names means that this particular matching works.

An angle is one letter. Each vertex names an angle, so the three angle pairs come straight from the three letter pairs:

\angle A \cong \angle D \qquad \angle B \cong \angle E \qquad \angle C \cong \angle F

When a vertex has several angles at it and a single letter would be unclear, use three letters and match each one: \angle BAC corresponds to \angle EDF, because B \leftrightarrow E, A \leftrightarrow D and C \leftrightarrow F.

A side is two letters. A side is named by its two endpoints, so its partner is named by the partners of those endpoints. Take the letters in positions one and two, two and three, and one and three:

\overline{AB} \cong \overline{DE} \qquad \overline{BC} \cong \overline{EF} \qquad \overline{AC} \cong \overline{DF}

The order of the two letters inside one side’s name does not matter, since \overline{FD} and \overline{DF} are the same segment. What matters is which two vertices are joined.

These six congruences are exactly what the statement claims. Put the other way round: corresponding parts of congruent triangles are congruent, which is the definition read backwards, and in proofs it becomes a reason of its own (Corresponding parts of congruent triangles).

Reorder both names the same way. The statement is about pairs of letters, not about which pair is written first. Rearranging both names in the same way keeps every pair, so

\begin{gathered} \triangle ABC \cong \triangle DEF \\ \triangle BCA \cong \triangle EFD \\ \triangle CBA \cong \triangle FED \end{gathered}

all say the same thing. So does \triangle DEF \cong \triangle ABC, with the two sides swapped, because congruence is symmetric (Properties of equality and congruence). What you may not do is reorder one name alone: that changes the pairs.

Write a statement from a marked figure. Name the first triangle in any order you like, then find each vertex’s partner and write it in the same position. Arcs point to vertices directly: the vertex with two arcs goes with the other vertex with two arcs. If only the sides are marked, use the fact that a vertex is where two sides meet: the vertex between the one-tick and two-tick sides matches the vertex, in the other triangle, between its one-tick and two-tick sides.

Mark a figure from a statement. Run the same reasoning the other way: list the six pairs, give each pair of sides its own number of ticks and each pair of angles its own number of arcs.

A part the two triangles share. Sometimes two triangles have a side in common, like the diagonal \overline{AC} of the kite in Figure 2.

Figure 2The diagonal \overline{AC} belongs to both triangles, so it is matched with itself.

The marks give B \leftrightarrow D (three arcs each), and A and C each match themselves, so

\triangle ABC \cong \triangle ADC

Reading off the sides, \overline{AB} \cong \overline{AD}, \overline{BC} \cong \overline{DC} and \overline{AC} \cong \overline{AC}. The last pair is a side matched with itself. It needs no mark on the figure, because every segment is congruent to itself (the reflexive property), and in a proof that is the reason you cite for it.

Worked examples

Example 1

Listing the six pairs

Given \triangle JKL \cong \triangle PQR, list all six pairs of congruent parts.

Solution

Match the letters by position: J \leftrightarrow P, K \leftrightarrow Q, L \leftrightarrow R.

Each vertex pair gives an angle pair:

\angle J \cong \angle P \qquad \angle K \cong \angle Q \qquad \angle L \cong \angle R

Each side is two vertices, so take positions one and two, two and three, one and three:

\overline{JK} \cong \overline{PQ} \qquad \overline{KL} \cong \overline{QR} \qquad \overline{JL} \cong \overline{PR}

Check. Every letter of \triangle JKL appears in exactly one angle pair and in exactly two side pairs, since each vertex is an endpoint of two sides. It does.

Example 2

Reading a statement from the marks

Write a congruence statement for the triangles in Figure 3, starting with \triangle PQR.

Figure 3Read the matching from the marks, not from the alphabet.

Solution

The name on the left is fixed, so find the partners of P, Q and R, in that order.

\angle P has one arc. The only other angle with one arc is at Z, so P \leftrightarrow Z. \angle Q has two arcs, and so does \angle X, so Q \leftrightarrow X. \angle R has three arcs, matching \angle Y, so R \leftrightarrow Y. Therefore

\triangle PQR \cong \triangle ZXY

Check with the sides. The statement claims \overline{PQ} \cong \overline{ZX}, \overline{QR} \cong \overline{XY} and \overline{PR} \cong \overline{ZY}. On the figure \overline{PQ} and \overline{ZX} carry one tick, \overline{QR} and \overline{XY} two, and \overline{RP} and \overline{YZ} three. All three agree.

The alphabetical \triangle PQR \cong \triangle XYZ would claim \angle P \cong \angle X, a one-arc angle with a two-arc angle, which the figure says is false.

Example 3

Which statements say the same thing?

Suppose \triangle ABC \cong \triangle DEF. Which of these say the same thing?

(a) \triangle BCA \cong \triangle EFD
(b) \triangle ACB \cong \triangle DEF
(c) \triangle FDE \cong \triangle CAB
(d) \triangle DEF \cong \triangle ABC
(e) \triangle CBA \cong \triangle FDE

Solution

Two statements say the same thing exactly when they pair the same letters. The original pairs are A with D, B with E, C with F. Read the pairs off each one.

(a) B with E, C with F, A with D: the same three pairs in a new order. Same.

(b) A with D, C with E, B with F. Only one pair survives. Different. It claims, for instance, \angle C \cong \angle E and \overline{AC} \cong \overline{DE}, neither of which the original says. Only one name was reordered, which is exactly the move that changes the pairs.

(d) The two names swapped whole. Same.

(e) C with F, B with D, A with E. Only C with F survives. Different.

So (a), (c) and (d) say the same thing as \triangle ABC \cong \triangle DEF; (b) and (e) do not.

Example 4

Using a statement to find measures

\triangle ABC \cong \triangle DEF. In \triangle ABC, m\angle A = 50^\circ and AC = 13. In \triangle DEF, m\angle E = 72^\circ and DF = 2x + 3. Find x and the three angle measures of \triangle DEF.

Solution

Find which parts correspond before doing any arithmetic. \overline{AC} joins the first and third letters, so its partner is \overline{DF}. Corresponding sides are congruent, so their lengths are equal:

\begin{aligned} 2x + 3 &= 13 \\ 2x &= 10 \\ x &= 5 \end{aligned}

For the angles, \angle D corresponds to \angle A, so m\angle D = 50^\circ. That gives two angles of \triangle DEF, and the angle sum gives the third:

m\angle F = 180^\circ - 50^\circ - 72^\circ = 58^\circ

So x = 5, and \triangle DEF has angles 50^\circ, 72^\circ and 58^\circ at D, E and F. The statement then gives m\angle B = 72^\circ and m\angle C = 58^\circ for free.

Example 5

Two triangles that meet at a point

In Figure 4, \triangle ABC \cong \triangle DEC. Given m\angle A = (4x + 6)^\circ, m\angle D = (6x - 14)^\circ and m\angle B = 71^\circ, find x, m\angle DCE and m\angle E.

Figure 4Two triangles meeting at one point. The angles at C are vertical angles.

Solution

The correspondence is A \leftrightarrow D, B \leftrightarrow E, C \leftrightarrow C: the point C belongs to both triangles and is matched with itself.

\angle A and \angle D correspond, so they are equal:

\begin{aligned} 4x + 6 &= 6x - 14 \\ 20 &= 2x \\ x &= 10 \end{aligned}

So m\angle A = 4(10) + 6 = 46^\circ, and m\angle D = 6(10) - 14 = 46^\circ as well.

Next, \angle B \leftrightarrow \angle E, so m\angle E = 71^\circ.

The angle of \triangle ABC at C is \angle ACB. By the angle sum,

m\angle ACB = 180^\circ - 46^\circ - 71^\circ = 63^\circ

At C there are four angles, so use three letters to say which one corresponds. Map each letter of \angle ACB: A \to D, C \to C, B \to E. The partner is \angle DCE, so m\angle DCE = 63^\circ.

Check. \angle ACB and \angle DCE sit opposite each other where \overline{AD} crosses \overline{BE}, so they are vertical angles and must be equal anyway. The statement and the vertical angles agree, and \triangle DEC has angles 46 + 71 + 63 = 180.

Example 6

A triangle matched with itself

What does \triangle ABC \cong \triangle ACB say about triangle ABC? What does \triangle ABC \cong \triangle BCA say?

Solution

The statement means what it always means, all six pairs match; the two names just happen to use the same three letters.

In \triangle ABC \cong \triangle ACB the correspondence is A \leftrightarrow A, B \leftrightarrow C, C \leftrightarrow B. The six pairs are

\overline{AB} \cong \overline{AC} \qquad \overline{BC} \cong \overline{CB} \qquad \overline{AC} \cong \overline{AB}

\angle A \cong \angle A \qquad \angle B \cong \angle C \qquad \angle C \cong \angle B

Two pairs say nothing new (\overline{BC} with itself, \angle A with itself) and the rest repeat each other. What is left is \overline{AB} \cong \overline{AC} and \angle B \cong \angle C: the triangle is isosceles, and the two angles opposite the equal sides are equal.

In \triangle ABC \cong \triangle BCA the correspondence is A \leftrightarrow B, B \leftrightarrow C, C \leftrightarrow A. The sides give \overline{AB} \cong \overline{BC}, \overline{BC} \cong \overline{CA} and \overline{AC} \cong \overline{BA}, and the angles give \angle A \cong \angle B \cong \angle C. The triangle is equilateral, and equiangular too.

Both answers come from the definition alone, but only because the statement was given. Whether a triangle with two equal sides, or two equal angles, really is congruent to itself in the order ACB is a different question, needing a way to prove triangles congruent; answering it is how the Base Angles Theorem and its converse are proved.

Boundaries

A wrongly ordered statement is false. Two triangles can be congruent while a particular statement about them is false, because the statement names a particular matching: \triangle PQR \cong \triangle XYZ is false for Figure 3 even though the triangles are congruent. “The triangles are congruent” and “\triangle PQR \cong \triangle ZXY” are different sentences, and only the second tells you which parts are equal.

Marks decide, not appearance. A figure need not be drawn to scale, and two angles that look equal are not known to be equal until something says so. Read the correspondence from the ticks, arcs and given facts, never from what looks right; see what a diagram may assume.

This lesson does not decide whether triangles are congruent. A congruence statement is justified by showing all six pairs match, or by a shorter list that is known to force the rest. Those shorter lists are SSS, SAS, ASA and AAS.

A symmetric triangle has more than one correct statement. For scalene triangles, with three different side lengths, only one matching works (Exercise 22). Two congruent isosceles triangles that are not equilateral can be matched two ways, the equal sides swapped; equilateral ones all six ways.

The same rule works for any polygon. ABCD \cong PQRS matches four vertices in order and claims four pairs of sides and four pairs of angles. Similar figures, same shape but not necessarily the same size, use the same letter order with the symbol \sim; see similar triangles, later.

Where this goes next

Once you can write a statement, the next question is how little you need to know to be sure it is true. SSS, SAS, ASA and AAS answer that: three well-chosen pairs force the other three. A proof then establishes a congruence by one of those criteria and spends it by reading off a pair it never measured, which is proof with corresponding parts.

The careful letter-matching of this lesson matters most when triangles overlap and share sides or angles, as in the kite of Figure 2. Those figures get a lesson of their own in proofs with overlapping triangles.

Exercises

Reading a statement

\triangle RST \cong \triangle UVW. List all six pairs of congruent parts.
\triangle MNP \cong \triangle XZY. Which angle corresponds to \angle N? Which side corresponds to \overline{MP}?
\triangle GHK \cong \triangle LMN. Which side corresponds to \overline{KH}? Which angle corresponds to \angle L?
\triangle ABC \cong \triangle FED. Fill in the blanks: \angle C \cong ___, \overline{BC} \cong ___, and ___ \cong \overline{FD}.

Same statement or not

\triangle ABC \cong \triangle DEF. Which of these say the same thing? (a) {\triangle CAB \cong \triangle FDE}, (b) {\triangle BAC \cong \triangle DEF}, (c) {\triangle EFD \cong \triangle BCA}, (d) {\triangle ACB \cong \triangle DFE}.
Rewrite \triangle PQR \cong \triangle STU so that the triangle on the right is named starting with U.
In how many different ways can the statement \triangle ABC \cong \triangle DEF be written? Explain the count.

Reading a figure

Use Figure 5 for Exercises 8 and 9, Figure 6 for Exercises 10 and 11, and Figure 2 for Exercise 12.

Figure 5Sides only. Each vertex is found from the two sides that meet there.

Figure 6Angles only. Each side is found from the two vertices at its ends.

Write a congruence statement for the triangles in Figure 5, starting with \triangle DEF.
In Figure 5, which angle is congruent to \angle E? Which angle is congruent to \angle S?
Write a congruence statement for the triangles in Figure 6, starting with \triangle GHJ.
In Figure 6, which side is congruent to \overline{GH}? Which side is congruent to \overline{LM}?
In Figure 2, write the congruence statement that starts with \triangle CBA, and name the angle that corresponds to \angle BAC.

Finding measures

\triangle ABC \cong \triangle DEF, with AB = 9, BC = 12 and AC = 14. Find DE, EF and DF, and the perimeter of \triangle DEF.
\triangle JKL \cong \triangle XYZ, with m\angle J = 38^\circ and m\angle Y = 85^\circ. Find m\angle K, m\angle L, m\angle X and m\angle Z.
\triangle PQR \cong \triangle STU, with PQ = 3x - 4 and ST = x + 10. Find x and PQ.
\triangle ABC \cong \triangle DEF, with m\angle B = (5y + 2)^\circ and m\angle E = (7y - 18)^\circ. Find y and m\angle B.
\triangle GHI \cong \triangle JKL, with GH = 2a + b, HI = a + 2b, JK = 11 and KL = 10. Find a, b, GH and HI.

Reasoning

What does \triangle ABC \cong \triangle BAC tell you about triangle ABC? Explain.
For the same pair of triangles, Mia writes \triangle ABC \cong \triangle DEF and Leo writes \triangle ABC \cong \triangle EDF. Can both be right? If so, what must be true of \triangle DEF?
Two triangles have the same perimeter, 12. Must they be congruent? Explain with an example.
\triangle ABC \cong \triangle DEF and \triangle DEF \cong \triangle GHI. Write a congruence statement relating \triangle ABC and \triangle GHI, and explain why each of its six pairs holds.
Explain why, for two congruent scalene triangles, only one correspondence works.

Answers

\angle R \cong \angle U, \angle S \cong \angle V, \angle T \cong \angle W, and \overline{RS} \cong \overline{UV}, \overline{ST} \cong \overline{VW}, \overline{RT} \cong \overline{UW}, matching by position.
\angle Z, and \overline{XY}. N and Z are both second; \overline{MP} joins the first and third letters, so its partner joins X and Y.
\overline{NM} (the same segment as \overline{MN}), and \angle G. K \leftrightarrow N, H \leftrightarrow M, and L is first, like G.
\angle C \cong \angle D, \overline{BC} \cong \overline{ED}, and \overline{AC} \cong \overline{FD}, from A \leftrightarrow F, B \leftrightarrow E, C \leftrightarrow D.
(a), (c) and (d): each pairs A with D, B with E and C with F. (b) pairs B with D and A with E, so it is different.
\triangle RPQ \cong \triangle UST (or \triangle RQP \cong \triangle UTS). U is the partner of R, so R comes first on the left, and the other letters keep their partners.
12 ways. The left name has 3 \times 2 \times 1 = 6 orders, each forcing one order on the right, and the two triangles can go either side of \cong, so 6 \times 2 = 12.
\triangle DEF \cong \triangle TRS. D and T each lie between the one-tick and three-tick sides, E and R between one and two, F and S between two and three.
\angle R is congruent to \angle E, and \angle F is congruent to \angle S, since E \leftrightarrow R and F \leftrightarrow S (Exercise 8).
\triangle GHJ \cong \triangle MKL. One arc at G and at M, two at H and at K, three at J and at L.
\overline{MK} is congruent to \overline{GH}, and \overline{JG} is congruent to \overline{LM}, from G \leftrightarrow M, H \leftrightarrow K, J \leftrightarrow L.
\triangle CBA \cong \triangle CDA, and \angle DAC. With C \leftrightarrow C, B \leftrightarrow D, A \leftrightarrow A, the letters of \angle BAC map to D, A, C.
DE = 9, EF = 12, DF = 14, and the perimeter is 9 + 12 + 14 = 35.
m\angle K = 85^\circ, m\angle L = 57^\circ, m\angle X = 38^\circ and m\angle Z = 57^\circ. \angle K matches \angle Y and \angle X matches \angle J; the angle sum gives m\angle L = 180^\circ - 38^\circ - 85^\circ = 57^\circ, and \angle Z matches \angle L.
x = 7 and PQ = 17. \overline{PQ} corresponds to \overline{ST}, so 3x - 4 = x + 10, giving 2x = 14. Check: 3(7) - 4 = 17 = 7 + 10.
y = 10 and m\angle B = 52^\circ. \angle B corresponds to \angle E, so 5y + 2 = 7y - 18, giving 20 = 2y. Check: 5(10) + 2 = 52 = 7(10) - 18.
a = 4, b = 3, GH = 11 and HI = 10. \overline{GH} matches \overline{JK} and \overline{HI} matches \overline{KL}, so 2a + b = 11 and a + 2b = 10. Subtracting the second from twice the first gives 3a = 12, so a = 4 and then b = 3.
\overline{BC} \cong \overline{AC} and \angle A \cong \angle B: the triangle is isosceles, its equal sides meeting at C. The correspondence is A \leftrightarrow B, B \leftrightarrow A, C \leftrightarrow C; the pair \overline{AB} with \overline{BA} says nothing, and the others give the two facts above.
Yes, but only if \overline{DF} \cong \overline{EF} and \angle D \cong \angle E, that is, \triangle DEF is isosceles with base \overline{DE}. Mia’s statement gives \overline{BC} \cong \overline{EF} and Leo’s gives \overline{BC} \cong \overline{DF}; likewise \angle A is congruent to both \angle D and \angle E. For scalene triangles at most one of them is right.
No. Triangles with sides 3, 4 and 5 and with sides 4, 4 and 4 both have perimeter 12, but the second has three equal sides and the first none, so no matching can pair every side with a side of the same length.
\triangle ABC \cong \triangle GHI, following each letter through: A \to D \to G, B \to E \to H, C \to F \to I. Each part of \triangle ABC is congruent to its partner in \triangle DEF, which is congruent to its partner in \triangle GHI; for example \overline{AB} \cong \overline{DE} and \overline{DE} \cong \overline{GH}, so \overline{AB} \cong \overline{GH} by the transitive property of congruence. The same holds for all six pairs.
The three sides have three different lengths, and congruent sides have equal lengths, so the shortest side must be matched with the shortest, the middle with the middle and the longest with the longest. That fixes the side matching, and since each vertex is where two particular sides meet, it fixes the vertex matching too.