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Using Congruence in Proofs: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 Extract one part from a proved match

A proof has established $\triangle KLM\cong\triangle PRQ$ . The given lengths are:
$\begin{aligned}KM&=2x+9\text{ cm} \\ PQ&=41\text{ cm}\end{aligned}.$
Find $x$ and explain which correspondence you used.
1. Hint 1
  Match the endpoints of KM using the ordered triangle names.
2. Hint 2
  Set the two corresponding length expressions equal.
Answer

$x=16$ ; KM corresponds to PQ.

Full solution

The ordered match sends $K$ to $P$ and $M$ to $Q$ .
By CPCTC, $KM=PQ$ .

Set the lengths equal:
$\begin{aligned}2x+9&=41\end{aligned}.$

Subtract 9 and divide by 2 to obtain:
$\begin{aligned}x&=16\end{aligned}.$

The corresponding lengths both equal $41$ cm.

Answer

$x=16$ ; KM corresponds to PQ.

Key idea

Use the established vertex order before translating a target into algebra.
Problem 2 Select the whole-triangle reason

Each case describes two actual triangles $ABC$ and $DEF$ in the stated order. Name a valid congruence criterion and the reason for concluding $\angle C\cong\angle F$ afterward.

Case A.
$\begin{aligned}AB&=DE \\ BC&=EF \\ AC&=DF\end{aligned}.$
Case B.
$\begin{aligned}AB&=DE \\ \angle BAC&\cong\angle EDF \\ \angle ABC&\cong\angle DEF\end{aligned}.$
Case C. Both triangles are right at $B$ and $E$ . Their given side pairs are:
$\begin{aligned}AC&=DF \\ AB&=DE\end{aligned}.$

Each case has its own given markings.

Text description of this figure

Case A marks three side pairs. Case B marks the included side and endpoint angles. Case C marks right angles, matching hypotenuses and one leg pair.
1. Hint 1
  Check which known parts are sides, angles, or the hypotenuse.
2. Hint 2
  Separate the reason for congruence from the reason for the final angle match.
Answer

Case A: SSS then CPCTC. Case B: ASA then CPCTC. Case C: HL then CPCTC.

Full solution

Case A.
The three side pairs establish $\triangle ABC\cong\triangle DEF$ by SSS.
Angle $C$ corresponds to angle $F$ , so CPCTC gives their congruence.

Case B.
The equal side is included between the two known angles.
ASA establishes the ordered triangle match; CPCTC then compares angles $C$ and $F$ .

Case C.
The side opposite each right angle is the hypotenuse.
The hypotenuses satisfy $AC=DF$ .
The given legs satisfy $AB=DE$ .
HL establishes congruence, followed by CPCTC for the target angles.

Answer

Case A: SSS then CPCTC. Case B: ASA then CPCTC. Case C: HL then CPCTC.

Key idea

A congruence criterion establishes the whole match; CPCTC establishes its target consequence.
Problem 3 Compute through an overlap

Points $A,B,D$ lie on one ray in that order, and $A,C,E$ lie on a second distinct ray in that order, with opening strictly between $0^\circ$ and $180^\circ$ . The given lengths are:
$\begin{aligned}AB&=AC \\ AC&=8\text{ cm} \\ AD&=AE \\ AE&=21\text{ cm} \\ BE&=3t+2\text{ cm} \\ CD&=5t-10\text{ cm}\end{aligned}.$
Find $t$ and the common length of $BE$ and $CD$ .

The given rays and point labels are drawn.

Text description of this figure

The named points lie on the two rays in their stated orders. The segments whose lengths appear in the problem are drawn without equality markings.
1. Hint 1
  Choose the two triangles containing BE and CD.
2. Hint 2
  The paired short and long sides include the same angle at A.
3. Hint 3
  After matching the triangles, equate the two target expressions.
Answer

$\begin{aligned}t&=6\\BE&=20\text{ cm}\\CD&=20\text{ cm}\end{aligned}.$

Full solution

Compare $\triangle ABE$ with $\triangle ACD$ .
The side pairs are:
$\begin{aligned}AB&=AC \\ AE&=AD\end{aligned}.$
Their included angles are the same opening between the two rays.

By SAS, $\triangle ABE\cong\triangle ACD$ .
By CPCTC, $BE=CD$ .

Set the target expressions equal:
$\begin{aligned}3t+2&=5t-10\end{aligned}.$

Rearranging gives:
$\begin{aligned}2t&=12\end{aligned}.$
Dividing by 2 gives:
$\begin{aligned}t&=6\end{aligned}.$
Substitute to obtain:
$\begin{aligned}3(6)+2&=20\end{aligned}.$

The other expression gives:
$\begin{aligned}5(6)-10&=20\end{aligned}.$
Both lengths are $20$ cm.

The 20 cm side labels appear after the congruence and algebra.

Answer

$\begin{aligned}t&=6\\BE&=20\text{ cm}\\CD&=20\text{ cm}\end{aligned}.$

Key idea

A separated mental picture can expose the common angle in overlapping triangles.
Problem 4 Locate the first unjustified step

The only givens are:
$\begin{aligned}AB&=DE \\ AC&=DF \\ \angle ABC&\cong\angle DEF\end{aligned}.$
Read the proposed argument below.

Line 1. By the givens:
$\begin{aligned}AB&=DE \\ AC&=DF\end{aligned}.$
Line 2. By the givens, $\angle ABC\cong\angle DEF$ .

Line 3. By SAS, $\triangle ABC\cong\triangle DEF$ .

Line 4. By CPCTC, $BC=EF$ .

Identify the first unjustified line and explain its defect. State which replacement angle datum, instead of the supplied angle pair, would make the SAS step valid.

Only the two side pairs and supplied angle pair are marked.

Text description of this figure

Two triangles have vertices A, B, C and D, E, F. Only the given side and angle markings are shown.
1. Hint 1
  SAS requires the angle between the two named sides.
2. Hint 2
  Find the vertex where AB and AC meet, and do the same for DE and DF.
Answer

Line 3 is the first unjustified line. The given angles are not included between the two side pairs. Replacing them with $\angle BAC\cong\angle EDF$ would make SAS valid.

Full solution

Lines 1 and 2 simply repeat the givens, so they are justified.

Line 3 invokes SAS, but $AB$ and $AC$ meet at $A$ , not at $B$ .
Likewise $DE$ and $DF$ meet at $D$ , not at $E$ .
The supplied angle pair is not the included pair.

Without an established congruence, Line 4 cannot use CPCTC.

The replacement given angle pair would be:
$\angle BAC\cong\angle EDF.$
With those angles, the two side pairs and their included angle would establish $\triangle ABC\cong\triangle DEF$ by SAS.
Line 4 would then follow.

In the repaired hypothesis, the angle pair is between the two given sides.

Answer

Line 3 is the first unjustified line. The given angles are not included between the two side pairs. Replacing them with $\angle BAC\cong\angle EDF$ would make SAS valid.

Key idea

Check the criterion’s exact hypotheses before accepting the CPCTC step.
Problem 5 Transfer a distance across a pond

Accessible points $J$ and $K$ are $26$ m apart. Point $L$ is across a pond. On the opposite side of line $JK$ , an accessible point $M$ is arranged so that:
$\begin{aligned}\angle LJK&\cong\angle MJK \\ \angle LKJ&\cong\angle MKJ\end{aligned}.$
The given lengths are:
$\begin{aligned}JM&=43\text{ m} \\ JL&=3u+7\text{ m}\end{aligned}.$
Find $JL$ and $u$ in the exact model. State the congruence criterion that permits the transfer.

The unknown distance retains its given expression.

Text description of this figure

L and M are on opposite sides of baseline JK. Only the given angle pairs are marked. JK is labeled 26 m, JM 43 m, and JL with the given expression.
1. Hint 1
  The same baseline lies between the two matched angle vertices.
2. Hint 2
  Find the accessible side corresponding to JL.
Answer

ASA followed by CPCTC gives the results:
$\begin{aligned}JL&=43\text{ m} \\ u&=12\end{aligned}.$

Full solution

The angle pairs at $J$ and $K$ match.
The included side satisfies $JK=JK$ .
By ASA, $\triangle LJK\cong\triangle MJK$ .

The corresponding sides $JL$ and $JM$ are equal by CPCTC.
Thus:
$\begin{aligned}JL=43\text{ m}\end{aligned}.$

The expression for the inaccessible length therefore satisfies:
$\begin{aligned}3u+7&=43\end{aligned}.$

Subtract 7 and divide by 3, giving:
$\begin{aligned}u&=12\end{aligned}.$

These are exact conclusions for the stated model.
Approximate field measurements would give an approximate distance.

Answer

ASA followed by CPCTC gives the results:
$\begin{aligned}JL&=43\text{ m} \\ u&=12\end{aligned}.$

Key idea

A measured accessible side can represent its inaccessible congruent partner.
Problem 6 Explain the copied opening

Task A. On paper, draw any source angle at $O$ with opening strictly between $0^\circ$ and $180^\circ$ , and a target ray from $P$ . Choose a convenient positive compass radius and copy the source angle with compass and straightedge, keeping the arcs. Label the source-circle marks $A,B$ , the target-baseline mark $D$ , and the chosen target intersection $E$ . Compare the source and target protractor readings as a check.

Task B. Consider a separate exact copy with the same labels and initial radius $5$ cm. Its source chord is:
$AB=4\text{ cm}.$
Justify the following angle congruence without assuming it:
$\angle AOB\cong\angle DPE.$
Task A's freely chosen angle need not have these numerical lengths.

This is the separate exact copy in Task B. The freely drawn angle in Task A may differ.

Text description of this figure

For Task B, source triangle OAB and target triangle PDE have common radius 5 centimeters and chord length 4 centimeters. One tick marks each radius; two ticks mark the transferred chords. No angle equality is marked.
1. Hint 1
  For Task A, recall which compass setting is kept when moving from the source vertex to the target vertex.
2. Hint 2
  The next setting is the distance between the two source-circle marks. Which two target circles must the new point lie on?
3. Hint 3
  For Task B, identify the three matching side pairs before drawing a conclusion about the two angles.
Answer

Task A: copy with arcs and matching protractor readings within drawing precision. Task B: SSS gives
$\triangle OAB\cong\triangle PDE.$
Then CPCTC gives
$\angle AOB\cong\angle DPE.$

Full solution

Task A.
Draw a circle of your chosen positive radius centered at $O$ to meet the source rays at $A$ and $B$ .
Keep that radius and draw a circle centered at $P$ , meeting the target ray at $D$ .

Task A.
Set the compass to chord $AB$ .
Draw an arc centered at $D$ to meet the target circle at $E$ on the desired side of the target baseline.
Draw ray $PE$ and retain the arcs.

Task A.
Measure the source and copied openings with a protractor.
Their readings should agree within drawing and measurement precision.
This checks the paper construction; the exact justification is the triangle argument.

Task B.
In the separate numerical copy, the source and target radii are all $5$ cm.
Thus:
$\begin{aligned}OA&=PD \\ OB&=PE\end{aligned}.$

Task B.
The transferred chord gives $AB=DE$ with both lengths $4$ cm.
The three equal side pairs establish $\triangle OAB\cong\triangle PDE$ .

Task B.
CPCTC gives $\angle AOB\cong\angle DPE$ .
The desired angle equality follows after SSS and was not an initial assumption.

The angle arcs belong to the conclusion, after the triangle match.

Answer

Task A: copy with arcs and matching protractor readings within drawing precision. Task B: SSS gives
$\triangle OAB\cong\triangle PDE.$
Then CPCTC gives
$\angle AOB\cong\angle DPE.$

Key idea

Copy the radius and chord to construct the angle, then use three side matches to justify its exact measure.
Problem 7 Choose a useful added segment

A four-sided figure $HJKL$ has every corner pointing outward. The given side pairs are:
$\begin{aligned}HJ&=HL \\ KJ&=KL\end{aligned}.$
Choose and draw a segment between opposite corners that divides the figure into two triangles guaranteed congruent by these givens. List the three matching side pairs, name the criterion, and identify the vertex corresponding to $J$ .

Matching ticks mark only the given side pairs.

Text description of this figure

H lies left, J above, K right, and L below. Matching ticks mark HJ with HL and KJ with KL. No diagonal is drawn.
1. Hint 1
  Look for two triangles whose sides can use both given equality pairs.
2. Hint 2
  Which segment between opposite corners would be shared by those two triangles?
Answer

Draw $HK$ . The SSS pairs and vertex match are
$\begin{aligned}HJ&=HL\\JK&=LK\\HK&=HK\\J&\leftrightarrow L\end{aligned}.$

Full solution

Draw the diagonal $HK$ , the segment joining the opposite corners $H$ and $K$ .
It creates $\triangle HJK$ and $\triangle HLK$ .

The first two side pairs are the givens:
$\begin{aligned}HJ&=HL \\ JK&=LK\end{aligned}.$

The third pair is the shared diagonal, so $HK=HK$ .

The three side matches give $\triangle HJK\cong\triangle HLK$ by SSS.
In that order, $J$ matches $L$ while $H$ and $K$ match themselves.
The requested corresponding vertex is $L$ .

The added diagonal gives the shared side of the two triangles.

Answer

Draw $HK$ . The SSS pairs and vertex match are
$\begin{aligned}HJ&=HL\\JK&=LK\\HK&=HK\\J&\leftrightarrow L\end{aligned}.$

Key idea

Drawing a useful diagonal exposes a reflexive side pair.
Problem 8 Use HL before solving for a leg

Triangles $ABC$ and $DEF$ are right at $B$ and $E$ . The given hypotenuse and leg pairs are:
$\begin{aligned}AC&=DF \\ DF&=50\text{ cm} \\ AB&=DE \\ DE&=14\text{ cm}\end{aligned}.$
Further measurements are:
$\begin{aligned}EF&=48\text{ cm} \\ BC&=5v+3\text{ cm}\end{aligned}.$
Find $v$ using a congruence argument.

Only the given lengths and right angles are labeled.

Text description of this figure

The two given right triangles have hypotenuse 50 cm and one leg 14 cm. The second remaining leg is labeled 48 cm; the first is labeled with its given expression.
1. Hint 1
  Identify the hypotenuses and the pair of given matching legs.
2. Hint 2
  After HL establishes the match, use the remaining leg pair.
Answer

$v=9$ .

Full solution

The two triangles are right triangles, the hypotenuses match, and one leg pair matches.
HL establishes $\triangle ABC\cong\triangle DEF$ .

The remaining legs correspond by CPCTC.
Therefore:
$\begin{aligned}BC&=EF \\ EF&=48\text{ cm}\end{aligned}.$
The resulting equation is:
$\begin{aligned}5v+3&=48\end{aligned}.$

Subtract 3 and divide by 5, giving:
$\begin{aligned}v&=9\end{aligned}.$

The first 48 cm label follows from HL and CPCTC.

Answer

$v=9$ .

Key idea

Use the right-angle condition with HL before transferring the other leg.
Problem 9 Read a crossing and compute an angle

Lines $AC$ and $BD$ cross at $O$ . The given distances are:
$\begin{aligned}OA&=OC \\ OC&=9\text{ cm} \\ OB&=OD \\ OD&=16\text{ cm}\end{aligned}.$
The given angle expressions are:
$\begin{aligned}m\angle OAB&=2w+5\text{ degrees} \\ m\angle OCD&=3w-15\text{ degrees}\end{aligned}.$
Find $w$ and the common angle measure, explaining why those two angles correspond.

The angle labels keep the unknown expressions.

Text description of this figure

Lines AC and BD cross at O. The given equal distance pairs are marked and labeled. The angles at A and C carry only their given algebraic expressions. No angle-equality marks are shown. Distinct arc patterns identify the two labeled angles without asserting equality.
1. Hint 1
  The two triangles use opposite rays at the crossing.
2. Hint 2
  Use the vertical angles with the two given length pairs, then read the vertex order.
Answer

$\begin{aligned}w&=20\\m\angle OAB&=45^\circ\\m\angle OCD&=45^\circ\end{aligned}.$

Full solution

Vertical angles give $\angle AOB\cong\angle COD$ .
The given length pairs are:
$\begin{aligned}OA&=OC \\ OB&=OD\end{aligned}.$
SAS establishes $\triangle AOB\cong\triangle COD$ .

The vertex match is:
$\begin{aligned}A&\leftrightarrow C \\ O&\leftrightarrow O \\ B&\leftrightarrow D\end{aligned}.$
By CPCTC, $\angle OAB\cong\angle OCD$ .

Set the measures equal:
$\begin{aligned}2w+5&=3w-15\end{aligned}.$

This gives:
$\begin{aligned}w&=20\end{aligned}.$
Substitute to get:
$\begin{aligned}2(20)+5&=45\end{aligned}.$

Each corresponding angle measures $45^\circ$ .

The numerical angle labels are results of the solution.

Answer

$\begin{aligned}w&=20\\m\angle OAB&=45^\circ\\m\angle OCD&=45^\circ\end{aligned}.$

Key idea

Vertical angles can supply the included pair needed before CPCTC applies.
Problem 10 Pass a result into a second congruence

Triangles $RAB$ and $SCD$ each have three noncollinear vertices. Their positive side lengths satisfy:
$\begin{aligned}RA&=SC \\ RB&=SD \\ AB&=CD\end{aligned}.$
First establish $\angle RAB\cong\angle SCD$ .

Point $E$ lies on ray $AR$ beyond $R$ , so $A,R,E$ occur in that order. Point $F$ lies on ray $CS$ beyond $S$ , so $C,S,F$ occur in that order. The additional given is:
$\begin{aligned}AE&=CF\end{aligned}.$
Prove:
$\begin{aligned}EB&=FD\end{aligned}.$
State the two ordered triangle matches and explain how the first result supports the second.

The rays retain the stated point orders and the given side markings.

Text description of this figure

A, R, E occur in that order on one ray, and C, S, F in that order on another. B and D lie off those rays. Only the stated side pairs RA/SC, RB/SD and AB/CD are marked. EB and FD have no matching marks.
1. Hint 1
  Match the first triangles from their three given side pairs, then identify the corresponding angle needed at A and C.
2. Hint 2
  Compare rays AR and AE, and compare rays CS and CF. What does each point order tell you?
3. Hint 3
  For the second comparison, combine the proved included-angle equality with the given side pairs.
Answer

SSS gives
$\triangle RAB\cong\triangle SCD.$
CPCTC gives
$\angle RAB\cong\angle SCD.$
SAS gives
$\triangle EAB\cong\triangle FCD.$
Then CPCTC gives
$EB=FD.$
The same rays connect the two angle pairs.

Full solution

The givens are:
$\begin{aligned}RA&=SC \\ RB&=SD \\ AB&=CD\end{aligned}.$
By SSS, $\triangle RAB\cong\triangle SCD$ .
The vertex match is:
$\begin{aligned}R&\leftrightarrow S \\ A&\leftrightarrow C \\ B&\leftrightarrow D\end{aligned}.$

CPCTC gives the corresponding angle equality $\angle RAB\cong\angle SCD$ .

Because $E$ lies on ray $AR$ , rays $AE$ and $AR$ coincide.
Thus $\angle EAB$ is $\angle RAB$ .
Similarly, rays $CF$ and $CS$ coincide, so $\angle FCD$ is $\angle SCD$ .
The included angles $\angle EAB$ and $\angle FCD$ are therefore congruent.

For the second pair, the given side pairs are:
$\begin{aligned}EA&=FC \\ AB&=CD\end{aligned}.$
With the proved included-angle pair, SAS establishes $\triangle EAB\cong\triangle FCD$ .
The vertex match is:
$\begin{aligned}E&\leftrightarrow F \\ A&\leftrightarrow C \\ B&\leftrightarrow D\end{aligned}.$

The remaining corresponding sides satisfy $EB=FD$ by CPCTC.

A proved angle pair, rather than a side pair, connects the two congruences.

Answer

SSS gives
$\triangle RAB\cong\triangle SCD.$
CPCTC gives
$\angle RAB\cong\angle SCD.$
SAS gives
$\triangle EAB\cong\triangle FCD.$
Then CPCTC gives
$EB=FD.$
The same rays connect the two angle pairs.

Key idea

A corresponding angle proved by one triangle match can supply the included angle for a second match.

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