Using Congruence in Proofs

Watch the video explainer

Learning goals

Plan congruence proofs with suitable triangles, a criterion, and an auxiliary segment when needed
Separate overlapping triangles and justify a shared side or shared angle
Pass a proved corresponding-part equality into a second triangle congruence
Use a congruent accessible triangle to determine an inaccessible distance
Justify the angle-copy construction using SSS followed by corresponding angles

Establish the triangles before using their parts

An exact triangle match lets us compare parts we did not use to establish it.

For example, suppose

\begin{aligned}AB&=DE \\ DE&=7\text{ cm} \\ AC&=DF \\ DF&=10\text{ cm} \\ m\angle BAC&=m\angle EDF \\ m\angle EDF&=52^\circ\end{aligned}.

The angle is included between the two named sides. Side-Angle-Side (SAS) establishes

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

The matching motions preserve angles, so the remaining corresponding angles agree too:

\begin{aligned}\angle ACB&\cong\angle DFE\end{aligned}.

That last step is called CPCTC, short for corresponding parts of congruent triangles are congruent. It uses a congruence already established; it does not establish triangle congruence by itself.

Worked example 1 Keep the two reasons separate

Problem. Given

\begin{aligned}JK&=MN \\ JL&=MP \\ \angle KJL&\cong\angle NMP\end{aligned}

prove

\begin{aligned}KL&=NP\end{aligned}.

Give a reason for the whole-triangle match and a reason for the target side equality.

The matching side ticks and included-angle arcs record the givens. The sides KL and NP are unmarked.

Solution. The angle at $J$ lies between $JK$ and $JL$ . Its partner at $M$ lies between $MN$ and $MP$ . The two side pairs and their included angle give

\begin{aligned}\triangle JKL&\cong\triangle MNP\end{aligned}

by SAS. The ordered match sends $K$ to $N$ and $L$ to $P$ . Therefore

\begin{aligned}KL&=NP\end{aligned}

by CPCTC.

Check your understanding

A proof has established triangle ABC congruent to triangle QPR. Which conclusion and reason are valid next?

A AC=QR by CPCTC B AC=PR by SSS C The triangles are congruent by CPCTC D AB=PR because they look alike

Work backward from a target: identify triangles containing its two parts, then ask which known criterion the givens can supply. Write the proof forward, using only justified statements. Keep the target equality out of the evidence used to establish congruence; otherwise the proof assumes its own conclusion.

The criteria themselves were justified earlier: Side-Side-Side (SSS), SAS and Angle-Side-Angle (ASA), and Angle-Angle-Side (AAS) and Hypotenuse-Leg (HL). Here the choice depends on the parts actually known, including whether an angle or side is included and whether both triangles are right triangles.

Shared and overlapping parts

A shared segment can match itself:

\begin{aligned}PR&=PR\end{aligned}

is the reflexive property of equality. An angle can also be shared, but its rays must really be the same rays. Sharing a vertex alone does not make two angles equal.

In a crowded figure, redraw the two triangles separately. Preserve every vertex name, given mark, and correspondence; do not introduce a fact because the new drawing looks regular.

Worked example 2 Separate two overlapping triangles

Problem. Points $A,B,D$ lie on one ray in that order. Points $A,C,E$ lie on a second ray in that order. The rays are distinct and form an angle between $0^\circ$ and $180^\circ$ . Given

\begin{aligned}AB&=AC \\ AC&=6\text{ cm} \\ AD&=AE \\ AE&=15\text{ cm}\end{aligned}

prove

\begin{aligned}BE&=CD\end{aligned}.

The triangles ABE and ACD overlap. B and D are on one ray; C and E are on the other.

Solution. In $\triangle ABE$ and $\triangle ACD$ , the side pairs are

\begin{aligned}AB&=AC \\ AE&=AD\end{aligned}.

The angle between them is the same opening at $A$ : ray $AB$ is ray $AD$ , and ray $AE$ is ray $AC$ .

Thus

\begin{aligned}\triangle ABE&\cong\triangle ACD\end{aligned}

by SAS, with

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

The remaining side pair gives

\begin{aligned}BE&=CD\end{aligned}

by CPCTC.

After SAS establishes the match, the highlighted sides BE and CD are corresponding parts and therefore equal.

Check your understanding

In the overlapping figure, why are angle BAE and angle CAD the same angle?

A Their vertices have the same letter, which is always enough B Their sides have the same lengths C AB and AD are the same ray, and AE and AC are the same ray D All overlapping triangles share every angle

Add a useful segment

An auxiliary segment is an extra segment drawn to help an argument. A diagonal joins two opposite corners of a four-sided figure. Drawing one creates triangles to compare, but does not by itself make any new lengths or angles equal.

Worked example 3 Use a common diagonal

Problem. A four-sided figure has vertices $P,Q,R,S$ in order, with every corner pointing outward. Given

\begin{aligned}PQ&=PS \\ RQ&=RS\end{aligned}

prove

\begin{aligned}\angle QPR&\cong\angle SPR\end{aligned}

by adding a segment if needed.

Solution. Draw diagonal $PR$ . The two triangles have

\begin{aligned}PQ&=PS \\ RQ&=RS\end{aligned}

by the givens, and

\begin{aligned}PR&=PR\end{aligned}

because it is shared. Therefore

\begin{aligned}\triangle PQR&\cong\triangle PSR\end{aligned}

by SSS. Under this order, $P$ matches $P$ , $Q$ matches $S$ , and $R$ matches $R$ . Hence

\begin{aligned}\angle QPR&\cong\angle SPR\end{aligned}

by CPCTC.

The dashed diagonal supplies a common side. The target angle equality is established by the proof, rather than marked as a given.

Pass a result to a second match

One triangle match can establish information needed in another. The second comparison needs its own evidence and its own vertex order.

Worked example 4 Carry an intermediate side equality forward

Problem. The named triangles each have three noncollinear vertices. Given

\begin{aligned}RA&=SC \\ SC&=7\text{ cm} \\ RB&=SD \\ SD&=12\text{ cm} \\ m\angle ARB&=m\angle CSD \\ m\angle CSD&=50^\circ\end{aligned}

first establish

\begin{aligned}AB&=CD\end{aligned}.

Also given

\begin{aligned}EA&=FC \\ FC&=8\text{ cm} \\ EB&=FD \\ FD&=11\text{ cm}\end{aligned}.

Use your first result to prove

\begin{aligned}\angle AEB&\cong\angle CFD\end{aligned}.

State both ordered triangle congruences and justify the steps between them.

All edge labels are given lengths. The unmarked shared sides belong to both triangles in their configuration.

Solution. Stage 1 compares $\triangle RAB$ and $\triangle SCD$ . The $7$ cm and $12$ cm side pairs include the $50^\circ$ angle pair. Thus

\begin{aligned}\triangle RAB&\cong\triangle SCD\end{aligned}

by SAS. Here

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

Consequently

\begin{aligned}AB&=CD\end{aligned}

by CPCTC.

Stage 2 compares $\triangle EAB$ and $\triangle FCD$ . Its known side pairs are

\begin{aligned}EA&=FC \\ FC&=8\text{ cm} \\ EB&=FD \\ FD&=11\text{ cm}\end{aligned}.

The first stage supplies the third pair,

\begin{aligned}AB&=CD\end{aligned}

\begin{aligned}\triangle EAB&\cong\triangle FCD\end{aligned}

by SSS. This comparison has a new first-vertex match:

$\begin{aligned}E&\leftrightarrow F\end{aligned}.$

The other two vertex matches remain:

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

Therefore

\begin{aligned}\angle AEB&\cong\angle CFD\end{aligned}

by CPCTC.

The equal side pair passed between the stages is a proved result. It is different from a shared segment equaling itself by the reflexive property.

The first match supplies a side pair required by the second. Each arrow follows a justified step.

Check your understanding

After the first match gives

\begin{aligned}AB&=CD\end{aligned}

by CPCTC,

\begin{aligned}EA&=FC \\ EB&=FD\end{aligned}

are also known. What establishes triangle EAB congruent to triangle FCD?

A CPCTC establishes the second triangle match directly B SSS, using the two givens and the proved side pair C The first match's SAS reason automatically applies to every triangle D An equal side pair means the triangles share one segment

Measure a distance through its partner

Worked example 5 Replace an inaccessible triangle with an accessible one

Problem. Points $B$ and $C$ are on accessible land, with

\begin{aligned}BC&=19\text{ m}\end{aligned}.

Point $A$ is across a pond. On the other side of line $BC$ , an accessible point $D$ is chosen so that

\begin{aligned}\angle DBC&\cong\angle ABC \\ \angle BCD&\cong\angle BCA\end{aligned}.

Measurement gives

\begin{aligned}BD&=24\text{ m}\end{aligned}.

Find $AB$ in this ideal model and justify the transfer.

Copy the angles at both ends of the same baseline. The lower triangle can be measured directly.

Solution. The matched angles are at $B$ and $C$ . Their included side is the same segment $BC$ . Thus

\begin{aligned}\triangle ABC&\cong\triangle DBC\end{aligned}

by ASA, with

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

By CPCTC,

\begin{aligned}AB&=DB\end{aligned}

so the requested distance is $24$ m. The shared side $BC$ supplies the side equality for ASA; the $24$ m accessible side is the measurement transferred afterward.

Real angle sightings and length measurements have error. A field result based on approximate measurements would be reported as about $24$ m; exact equality belongs to the ideal geometric conditions in the problem.

Check your understanding

In the pond model, which side corresponds to AC after triangle ABC is matched with triangle DBC?

A BD B BC C AB D DC

Finish the angle-copy proof

In Circles and Construction Tools, the compass procedure copied distances, and the claim about the resulting angle was deferred. We can now prove it without assuming that the copied angle is equal.

Recall the labels. A circle centered at source vertex $O$ meets its two rays at $A$ and $B$ . A circle with the same radius, centered at target vertex $P$ , meets the target ray at $D$ . An arc centered at $D$ with radius $AB$ meets the target circle at $E$ , on the chosen side of the baseline. Draw ray $PE$ . The source opening is strictly between $0^\circ$ and $180^\circ$ , and the initial radius is positive, as in the original construction.

The radius setting transfers OA and OB as PD and PE. The chord setting transfers AB as DE. No angle equality is assumed.

Why the angle-copy construction works#

For a numerical instance, suppose the initial compass radius is $9$ cm and the chord $AB$ is $10$ cm. The source triangle has

\begin{aligned}OA&=OB \\ OB&=9\text{ cm} \\ AB&=10\text{ cm}\end{aligned}.

The target triangle has

\begin{aligned}PD&=PE \\ PE&=9\text{ cm} \\ DE&=10\text{ cm}\end{aligned}.

These are three side matches, so

\begin{aligned}\triangle OAB&\cong\triangle PDE\end{aligned}

by SSS. CPCTC gives

\begin{aligned}\angle AOB&\cong\angle DPE\end{aligned}.

For any permitted source angle and positive radius, the circle definitions give

\begin{aligned}OA&=PD \\ OB&=PE\end{aligned}.

The transferred chord gives

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

Therefore the same ordered triangles are congruent by SSS, regardless of the particular radius or chord length. Their corresponding angles at $O$ and $P$ agree by CPCTC. The source angle’s rays are $OA$ and $OB$ , and the constructed target rays are $PD$ and $PE$ , so these are precisely the angles being copied.

The existence of the required arc intersections is the accepted construction fact from Circles and Construction Tools. The three length equalities follow from the circle definitions and compass settings; the angle equality follows only after SSS.

Check your understanding

In the angle-copy construction, why is SSS used before concluding that the source and target angles agree?

A The compass transfers three side lengths, and the angles become corresponding parts of the resulting congruent triangles B The desired angle equality is an initial given for SAS C Every pair of angles drawn with a compass is equal D A chord has the same length as its central angle

The argument also completes the justification of using that angle copy in the parallel-line construction from Proving Lines Parallel or Perpendicular. There the corresponding-angle converse supplies parallelism once the copied angle is established.

Using Congruence in Proofs

Learning goals

Establish the triangles before using their parts

Shared and overlapping parts

Add a useful segment

Pass a result to a second match

Measure a distance through its partner

Finish the angle-copy proof

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice

Your progress