9.03 Using triangle congruence
Once we have shown two triangles are congruent using some of the side and angle information, we know the other sides and angles must match up as well.
Corresponding sides and angles in congruent triangles
In two congruent triangles, any sides or angles that match up are referred to as corresponding.
If two triangles are congruent, then:
- The sides in the same relative position are equal, and are called corresponding sides (of congruent triangles).
- The angles in the same relative position are equal, and are called corresponding angles (of congruent triangles).
Worked examples
Example 1
We can use the congruence test SSS to establish that $\triangle CDE\equiv\triangle LMN$△CDE≡△LMN.
What are the three corresponding angle pairs?
Think: If we arranged the two triangles so they were lying on top of each other, then $C$C and $L$L would be the same point. Similarly, $D$D and $M$M would be the same point, and $E$E and $N$N as well.
Do: Starting with $\angle DEC$∠DEC, we replace $D\rightarrow M$D→M, $E\rightarrow N$E→N, and $C\rightarrow L$C→L, which gives us $\angle MNL$∠MNL. Since these two angles have the same position, they are corresponding (and must be equal).
Doing this for each angle in $\triangle CDE$△CDE matches them with an angle in $\triangle LMN$△LMN:
$\angle DEC=\angle MNL$∠DEC=∠MNL
$\angle CDE=\angle LMN$∠CDE=∠LMN
$\angle ECD=\angle NLM$∠ECD=∠NLM
Example 2
We can use the congruence test AAS to establish that $\triangle PRQ\equiv\triangle SUT$△PRQ≡△SUT.
Which side in $\triangle STU$△STU is equal in length to $PQ$PQ?
Think: If we arranged the two triangles so they were lying on top of each other, then $P$P and $S$S would be the same point, and $Q$Q and $T$T would be the same as well.
Do: Starting with the side $PQ$PQ, we replace $P\rightarrow S$P→S and $Q\rightarrow T$Q→T to find the corresponding side $ST$ST. Since the triangles are congruent, these sides must be equal.
In both of these examples the order of the points making up each triangle was listed in the same order.
Example 1: $\triangle CDE\equiv\triangle LMN$△CDE≡△LMN means $C\leftrightarrow L$C↔L, $D\leftrightarrow M$D↔M, $E\leftrightarrow N$E↔N.
Example 2: $\triangle PRQ\equiv\triangle SUT$△PRQ≡△SUT means $P\leftrightarrow S$P↔S, $R\leftrightarrow U$R↔U, $Q\leftrightarrow T$Q↔T.
This is a very useful habit to develop when you are writing your own congruence statements, since it makes identifying corresponding sides and angles much easier.
Practice questions
Question 1
It is known that $\triangle STU\equiv\triangle ABC$△STU≡△ABC.
- Which two of the following equalities do we know to be true?
$TU=BC$TU=BC
A$US=AB$US=AB
B$TU=CA$TU=CA
C$ST=AB$ST=AB
D
Question 2
It is known that $\triangle STU\equiv\triangle PQR$△STU≡△PQR.
- Which two of the following equalities do we know to be true?
$\angle STU=\angle PQR$∠STU=∠PQR
A$\angle STU=\angle QPR$∠STU=∠QPR
B$SU=PQ$SU=PQ
C$ST=PQ$ST=PQ
D
Matching information across congruent triangles
If two corresponding sides or angles must be equal in congruent triangles then knowing the value of one gives us the value of the other.
Worked example
Example 3
We can use the congruence test AAS to establish that $\triangle GHI\equiv\triangle LMN$△GHI≡△LMN.
Find the value of $y$y.
Think: The side of length $y$y is opposite the angle of size $28^\circ$28°. The corresponding side in $\triangle GHI$△GHI must also have a length of $y$y.
Do: In $\triangle GHI$△GHI, the side $HI$HI is opposite the angle of size $28^\circ$28°, and it has length $6$6. This means $y=6$y=6 as well.
Practice questions
Question 3
These two triangles are congruent. Find the value of $y$y.
Two right-angled triangles, $\triangle STU$△STU and $\triangle ABC$△ABC are congruent. In $\triangle STU$△STU, the angle at vertex $U$U, $\angle TUS$∠TUS, is a right angle marked with a purple-shaded square. The angle at vertex $T$T, $\angle STU$∠STU, measures $44^\circ$44°. The angle at vertex $S$S, $\angle UST$∠UST, is unlabeled. The hypotenuse $ST$ST is opposite $\angle TUS$∠TUS at vertex $U$U. Side $SU$SU is opposite $\angle STU$∠STU at vertex $T$T. Side $TU$TU is opposite $\angle UST$∠UST at vertex $S$S.
In $\triangle ABC$△ABC, the angle at vertex $C$C, $\angle BCA$∠BCA, is a right angle marked with a purple-shaded square. The angle at vertex $B$B, $\angle ABC$∠ABC, is unlabeled. The angle at vertex $A$A, $\angle CAB$∠CAB, is labeled $y$y degrees. The hypotenuse $AB$AB is opposite $\angle BCA$∠BCA at vertex $C$C. Side $AC$AC is opposite $\angle ABC$∠ABC at vertex $B$B. Side $BC$BC is opposite $\angle CAB$∠CAB at vertex $A$A.
Both side $SU$SU of $\triangle STU$△STU and side $AC$AC of $\triangle ABC$△ABC are marked with single tick marks. Both side $ST$ST of $\triangle STU$△STU and side $AB$AB of $\triangle ABC$△ABC are marked double tick marks.
Both $\angle TUS$∠TUS at vertex $U$U of $\triangle STU$△STU and $\angle BCA$∠BCA at vertex $C$C of $\triangle ABC$△ABC are opposite the sides marked with double tick marks. Both $\angle STU$∠STU at vertex $T$T of $\triangle STU$△STU and $\angle ABC$∠ABC at vertex $B$B of $\triangle ABC$△ABC are opposite the sides marked with single tick marks. Both $\angle UST$∠UST of $\triangle STU$△STU and $\angle CAB$∠CAB of $\triangle ABC$△ABC are between the two marked sides.
Question 4
Consider the two triangles below:
Two isosceles triangles, $\triangle EFG$△EFG and $\triangle HJK$△HJK, have their vertices marked with solid dots.
The triangle above, $\triangle HJK$△HJK, has its vertices labeled $H$H, $J$J and $K$K arranged in counterclockwise order. The angle at vertex $H$H, $\angle JHK$∠JHK, measures $67^\circ$67° and is marked with a red-shaded double arc. The angle at vertex $J$J, $\angle HJK$∠HJK, measures $67^\circ$67° and is marked with a red-shaded double arc. Side $JK$JK, labeled as $m$m, is opposite $\angle JHK$∠JHK at vertex $H$H. Side $HK$HK is opposite $\angle HJK$∠HJK at vertex $J$J. Side $HJ$HJ is opposite $\angle HKJ$∠HKJ at vertex $K$K.
The triangle below, $\triangle EFG$△EFG, has its vertices labeled $E$E, $F$F, and $G$G arranged in counterclockwise. The angle at vertex $E$E, $\angle GEF$∠GEF, measures $67^\circ$67° and is marked with a red-shaded double arc. The angle at vertex $F$F, $\angle EFG$∠EFG, measures $67^\circ$67° and is marked with a red-shaded double arc. Side $EF$EF is opposite $\angle EGF$∠EGF at vertex $G$G. Side $EG$EG is opposite $\angle EFG$∠EFG at vertex $F$F. Side $FG$FG is opposite $\angle FEG$∠FEG at vertex $E$E.
Both side $HJ$HJ of $\triangle HJK$△HJK and side $EF$EF of $\triangle EFG$△EFG are between the marked angles. The sides, $JK$JK and $HK$HK of $\triangle HJK$△HJK and $EG$EG and $FG$FG of $\triangle EFG$△EFG, are all opposite the $67^\circ$67° angles.
Together with the given information, which other condition would make sure that these two triangles are congruent?
$\angle FGE=\angle JKH$∠FGE=∠JKH
A$EG=HJ$EG=HJ
B$EF=HJ$EF=HJ
C$EF=HK$EF=HK
DGiven that $EF=HJ$EF=HJ, and $EG=7$EG=7, find the value of $m$m.