Recall that when two triangles are congruent, one triangle can be transformed onto the other by a series of flips, slides, or turns.
The parts that match up from that transformation are called the corresponding parts. We know from the properties of transformations that the corresponding parts in congruent triangles are congruent.
In a congruence statement, the order of the triangle vertices tells us important information about the corresponding parts. Consider the following congruence statement:
$\Delta ABC\cong\Delta DEF$ΔABC≅ΔDEF
From the congruence statement above, we can conclude the following $6$6 pieces of information:
$\angle A\cong\angle D$∠A≅∠D | $\overline{AB}\cong\overline{DE}$AB≅DE |
---|---|
$\angle B\cong\angle E$∠B≅∠E | $\overline{BC}\cong\overline{EF}$BC≅EF |
$\angle C\cong\angle F$∠C≅∠F | $\overline{AC}\cong\overline{DF}$AC≅DF |
Given that the triangles $\Delta ABM$ΔABM and $\Delta CMD$ΔCMD are congruent, solve for $x$x:
Think: Which parts of each triangle are congruent? Using the congruence statement, we know that $\overline{AM}\cong\overline{CD}$AM≅CD. That also means that the sides have equal lengths.
Do: Write an equation for the lengths of the congruent sides, and substitute their values.
$AM$AM | $=$= | $CD$CD |
$x$x | $=$= | $7$7 |
Explain why $\Delta TUV$ΔTUV and $\Delta DFE$ΔDFE are congruent and solve for $x$x:
Think: Since we have been given two sides and an angle between them that do not have variables, we can say the triangles are congruent by SAS. We also know that congruent triangles have congruent corresponding parts. Our equation will be formed by equating measures of $\angle UTV$∠UTV and $\angle FDE$∠FDE.
Do: Write an equation relating the angle with the unknown measure and a known measure. In this case, we write the following equation, substitute, and then simplify:
$m\angle UTV$m∠UTV | $=$= | $m\angle FDE$m∠FDE |
$2x+20$2x+20 | $=$= | $34$34 |
$2x$2x | $=$= | $34-20$34−20 |
$2x$2x | $=$= | $14$14 |
$x$x | $=$= | $7$7 |
Reflect: Does the answer $x=7$x=7 make sense in the context of the diagram? How might you check your answer?
Consider the diagram below.
A triangle on the right, $\triangle ABC$△ABC, has vertices labeled $A$A, $B$B, and $C$C, and a triangle on the left, $\triangle EFD$△EFD, has vertices labeled $D$D, $E$E, and $F$F. The angle at vertex $A$A of $\triangle ABC$△ABC, $\angle BAC$∠BAC, is marked with a blue arc. The angle at vertex $E$E of $\triangle EFD$△EFD, $\angle DEF$∠DEF, is marked with an identical blue arc. Sides $interval(AB)$interval(AB) of $\triangle ABC$△ABC and $interval(EF)$interval(EF) of $\triangle EFD$△EFD have single hashmarks. Sides $interval(AC)$interval(AC) of $\triangle ABC$△ABC and $interval(DE)$interval(DE) of $\triangle EFD$△EFD have double hashmarks.In $\triangle ABC$△ABC, $\angle BAC$∠BAC is the included angle of sided $AB$AB and $AC$AC. In $\triangle EFD$△EFD, $\angle DEF$∠DEF is the included angle of sides $EF$EF and $DE$DE.
In $\triangle ABC$△ABC, $interval(AB)$interval(AB) is opposite to vertex $C$C, $interval(AC)$interval(AC) is opposite to vertex $B$B, and $interval(BC)$interval(BC) is opposite to vertex $A$A. In $\triangle EFD$△EFD, $interval(EF)$interval(EF) is opposite to vertex $D$D, $interval(DE)$interval(DE) is opposite to vertex $F$F, and $interval(DF)$interval(DF) is opposite to vertex $E$E.
Which of the following is a correct congruence statement for the triangles?
$\Delta ABC\cong\Delta EDF$ΔABC≅ΔEDF
$\Delta ABC\cong\Delta DEF$ΔABC≅ΔDEF
$\Delta ABC\cong\Delta EFD$ΔABC≅ΔEFD
State the reason why these two triangles are congruent.
SSS: All three corresponding sides are congruent.
HL: Two right triangles with hypotenuse and one leg are congruent.
SAS: A pair of corresponding sides and the included angle are congruent.
AAS: A pair of corresponding angles and a non-included side are congruent.
Which angle is congruent to $\angle ACB$∠ACB?
$\angle FED$∠FED
$\angle DFE$∠DFE
$\angle EDF$∠EDF
Consider the diagram below.
Find the value of $y$y.
Consider the adjacent figure:
Triangles $\triangle ACE$△ACE and $\triangle ABD$△ABD overlap such that they share vertex $A$A. $\triangle ACE$△ACE has sides $AC$AC, $AE$AE and $EC$EC. $\triangle ABD$△ABD has sides $AB$AB, $AD$AD and $BD$BD. Vertex $B$B of $\triangle ABD$△ABD lies along side $AC$AC of $\triangle ACE$△ACE, and vertex $E$E of $\triangle ACE$△ACE lies along side $AD$AD of $\triangle ABD$△ABD. Sides $BD$BD of $\triangle ABD$△ABD and $CE$CE of $\triangle ACE$△ACE intersect at point $F$F. The angle at vertex $A$A, which is both the $\angle CAE$∠CAE of $\triangle ACE$△ACE and $\angle DAB$∠DAB of $\triangle ABD$△ABD, is labeled and measures $45^\circ$45°.
Vertex $B$B divides side $AC$AC of $\triangle ACE$△ACE into two labeled segments: $AB$AB = $16$16 units and $BC$BC = $15$15 units. With vertex $E$E divides side $AD$AD of $\triangle ABD$△ABD into two labeled segments: $AE$AE = $16$16 units and $ED$ED = $15$15 units.
The exterior angle at vertex $E$E of $\triangle ACE$△ACE measures $82^\circ$82°. The $82^\circ$82° angle is formed by side $CE$CE of $\triangle ACE$△ACE and side $AD$AD of $\triangle ABD$△ABD, $\angle CED$∠CED, measures. The angle at vertex $D$D of $\triangle ABD$△ABD, $\angle ADB$∠ADB, is labeled $p$p-degrees.
Why are the two triangles $\Delta ACE$ΔACE and $\Delta ADB$ΔADB congruent?
Side-side-side congruence (SSS)
Angle-side-angle congruence (ASA)
Side-angle-side congruence (SAS)
Angle-angle-side congruence (AAS)
Now, solve for $p$p.
Recall that when two triangles are congruent, then their corresponding parts are congruent as well. You may have used this property to solve for missing values in congruent triangles.
Although it wasn't mentioned previously, this is a theorem that can be proven using the properties of transformations.
Corresponding parts of congruent triangles are congruent (CPCTC).
Or,
If two triangles are congruent, then their corresponding sides and angles are congruent.
Once we've proven this theorem is true, we can apply it in other proofs to justify other properties associated with triangles, quadrilaterals, and beyond!
Consider the following diagram:
Show that $\overline{AB}$AB is congruent to $\overline{CD}$CD.
Think: We have two triangles, and we need to show they are congruent first. The given information tells us one pair of congruent sides, one pair of congruent angles, and one pair of sides that both triangles share. We can use side-angle-side (SAS) to show these two triangles are congruent. Then we can conclude that all the sides of both triangles (including the two we are interested in!) have a matching congruent side in the other.
Do: Write what we want to show at the top, enter the given information as lines, and then use the correct steps and correct reasons to arrive at the conclusion.
To prove: $\overline{AB}\cong\overline{CD}$AB≅CD | |
Statement | Reason |
$\overline{AD}\cong\overline{CB}$AD≅CB | Given |
$\overline{BD}\cong\overline{DB}$BD≅DB | Reflexive property of congruence |
$\angle ADB\cong\angle CBD$∠ADB≅∠CBD | Given |
$\Delta ADB\cong\Delta CBD$ΔADB≅ΔCBD | Side-angle-side (SAS) congruence |
$\overline{AB}\cong\overline{CD}$AB≅CD | Corresponding parts of congruent triangles are congruent (CPCTC) |
Reflect: What else can we say about the line segments $\overline{AB}$AB and $\overline{CD}$CD? Are they parallel? Justify your reasoning.
This two-column proof shows that $\angle DEH\cong\angle FEG$∠DEH≅∠FEG in the attached diagram, but it is incomplete.
Statements | Reasons |
---|---|
$E$E is the midpoint of $\overline{DF}$DF | Given |
$\overline{DH}\cong\overline{FG}$DH≅FG | Given |
$\overline{EH}\cong\overline{EG}$EH≅EG |
Given |
$\left[\text{___}\right]$[___] | $\left[\text{___}\right]$[___] |
$\Delta DEH\cong\Delta FEG$ΔDEH≅ΔFEG |
Side-side-side congruence theorem (SSS) |
$\angle DEH\cong\angle FEG$∠DEH≅∠FEG | $\left[\text{___}\right]$[___] |
Select the correct pair of reasons to complete the proof.
$\overline{DE}\cong\overline{EF}$DE≅EF | Properties of a midpoint |
and | |
$\angle DEH\cong\angle FEG$∠DEH≅∠FEG | Side-side-side congruence theorem (SSS) |
$\angle DEH\cong\angle FEG$∠DEH≅∠FEG | Corresponding parts of congruent triangles are congruent (CPCTC) |
and | |
$\angle DEH\cong\angle FEG$∠DEH≅∠FEG | Side-side-side congruence theorem (SSS) |
$\overline{DE}\cong\overline{EF}$DE≅EF | Corresponding parts of congruent triangles are congruent (CPCTC) |
and | |
$\angle DEH\cong\angle FEG$∠DEH≅∠FEG | Properties of a midpoint |
$\overline{DE}\cong\overline{EF}$DE≅EF | Properties of a midpoint |
and | |
$\angle DEH\cong\angle FEG$∠DEH≅∠FEG | Corresponding parts of congruent triangles are congruent (CPCTC) |
This two-column proof shows that $\Delta RMP$ΔRMP is isosceles in the attached diagram, but it is incomplete.
Statements | Reasons |
---|---|
$\overline{RQ}\cong\overline{QP}$RQ≅QP | Given |
$\overline{RP}\perp\overline{MQ}$RP⊥MQ | Given |
$\angle MQR$∠MQRand$\angle MQP$∠MQP are right angles |
Definition of perpendicular |
$\angle MQR\cong\angle MQP$∠MQR≅∠MQP | All right angles are congruent |
$\overline{MQ}\cong\overline{MQ}$MQ≅MQ | Reflexive property of congruence |
$\left[\text{___}\right]$[___] |
$\left[\text{___}\right]$[___] |
$\left[\text{___}\right]$[___] | $\left[\text{___}\right]$[___] |
$\Delta RMP$ΔRMP is isosceles | Isosceles triangles are triangles with two congruent sides |
Triangle $\triangle MPR$△MPR has an inscribed line $MQ$MQ perpendicular to its side $RP$RP. Point $Q$Q is along the line $RP$RP. The angle formed by $QP$QP and $MQ$MQ is marked by a square to indicate a right angle. Sides $RQ$RQ and $QP$QP are marked with single tick, indicating congruence.
In triangle(MRQ),angle $\angle MQR$∠MQR is an included angle of sides $RQ$RQ and $MQ$MQ . And in triangle(MPQ), angle $\angle MQR$∠MQR is an included angle of sides $PQ$PQ and $MQ$MQ .
Select the correct pair of reasons to complete the proof.
$\overline{RM}\cong\overline{PM}$RM≅PM | Corresponding parts of congruent triangles are congruent (CPCTC) |
followed by | |
$\Delta RMQ\cong\Delta PMQ$ΔRMQ≅ΔPMQ | Side-angle-side congruence (SAS) |
$\Delta RMQ\cong\Delta PMQ$ΔRMQ≅ΔPMQ | Side-angle-side congruence (SAS) |
followed by | |
$\overline{RM}\cong\overline{PM}$RM≅PM | Corresponding parts of congruent triangles are congruent (CPCTC) |
$\Delta RMQ\cong\Delta PMQ$ΔRMQ≅ΔPMQ | Angle-side-angle congruence (ASA) |
followed by | |
$\overline{RM}\cong\overline{PM}$RM≅PM | Corresponding parts of congruent triangles are congruent (CPCTC) |
$\overline{RM}\cong\overline{PM}$RM≅PM | Corresponding parts of congruent triangles are congruent (CPCTC) |
followed by | |
$\Delta RMQ\cong\Delta PMQ$ΔRMQ≅ΔPMQ | Angle-side-angle congruence (ASA) |