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9.05 Congruence transformations

Lesson

Just as we can apply the properties of equality to real numbers to solve for an equation, we can also apply them to real number measurements when solving for values in a geometric diagram.

You may have been doing this already without even knowing it!

Properties of equality for real numbers

What exactly are the properties of equality?  Put simply, they're assumptions of how we can manipulate equations while maintaining truth.  They're the foundation for most of mathematics. 

The table below summarizes the properties and gives an example of how we might apply them in geometry using segment lengths.

Property of equality Meaning Example in geometry
Reflexive property Anything is equal to itself. $AB=AB$AB=AB
Symmetric property The reverse equation is also true. If $AB=CD$AB=CD then $CD=AB$CD=AB.
Transitive property Equality is transferable. If $AB=CD$AB=CD and $CD=EF$CD=EF, then $AB=EF$AB=EF.
Addition / Subtraction property Equals added to (or subtracted from) equals are still equal. If $AB=CD$AB=CD then $AB+5=CD+5$AB+5=CD+5
Multiplication / Division property Equals multiplied (or divided) by equals are still equal, as long as the division isn't by $0$0. If $AB=CD$AB=CD then $\frac{AB}{5}=\frac{CD}{5}$AB5=CD5

Since these properties are true for any real number, they're also true for real number measurements, such as segment lengths, or angle measures.

 

Properties of Congruence

We've established that measurements of geometrical objects follow the same properties of equality that real numbers do.  What can we say about the geometrical objects themselves?

We say objects are congruent if and only if one can be transformed into the other by a series of translations, rotation, and reflections.  We'll use this definition to prove that the reflexive, symmetric, and transitive properties of congruence are true.

Since we can perform a series of translations, rotations, and reflections that map an object onto itself (or bring it right back to where it started), the reflexive property holds for congruence.

If we can perform a series of translations, rotations, and reflections that bring object $A$A to object $B$B, then we can perform the reverse set of transformations to bring object $B$B to object $A$A.  This means that the order in a congruence statement doesn't matter, and the symmetric property of congruence holds.

Now suppose that object $A$A is congruent to object $B$B.  This means there is a series of transformations that map $A$A to $B$B.  Now suppose that object $B$B is congruent to object $C$C.  Then there's another series of transformations that map $B$B to $C$C.  If we combine the two series of transformations, then we can map $A$A to $C$C.  This means that $A$A and $C$C are also congruent.  Therefore, the transitive property of congruence holds.

The table below summarizes the properties of congruence with some statements using angles as examples.

Property of congruence Meaning Example in geometry
Reflexive property Anything is congruent to itself. $\angle ABC\cong\angle ABC$ABCABC
Symmetric property The reverse congruence statement is also true. If $\angle ABC\cong\angle DEF$ABCDEF then $\angle DEF\cong\angle ABC$DEFABC.
Transitive property Congruence is transferable. If $\angle ABC\cong\angle DEF$ABCDEF and $\angle DEF\cong\angle GHI$DEFGHI, then $\angle ABC\cong\angle GHI$ABCGHI.

Now that we've established that the above properties of congruence are true, we can apply them to solve problems.  Let's try some examples together.

Practice questions

Question 1

Consider the true expression:

If $AB=CD$AB=CD then $CD=AB$CD=AB

Which of the following properties does this illustrate?

  1. The transitive property of equality.

    A

    The multiplication property of equality.

    B

    The reflexive property of equality.

    C

    The addition property of equality.

    D

    The symmetric property of equality.

    E

Question 2

Fill in the blank so that the resulting statement is true.

  1. The symmetric property of equality states that if $AB=CD$AB=CD then $CD$CD = $\editable{}$.

Question 3

Consider the following true statement:

$AB+CD=32$AB+CD=32

  1. Which of the following is a valid deduction from this statement using only a single property of equality?

    $BA+DC=16+16$BA+DC=16+16

    A

    $AB+AB=32+CD$AB+AB=32+CD

    B

    $2AB=64-CD$2AB=64CD

    C

    $32=AB+CD$32=AB+CD

    D

We say that two triangles are congruent if one can be moved, rotated, and possibly reflected to lie on top of each other exactly. You can think of congruence as a more precise way of saying that two triangles are "the same". Here is an example to illustrate:

Here we are told by the markings on each triangle that the three sides are equal, and the three angles are equal as well - this is the given information. With a reflection, a rotation, and some translation, we will be able to place these triangles directly on top of each other.

Most of the time we are not given all three sides and all three angles. So the big question is: what is the minimum amount of information we need to know about two triangles to conclude that they are congruent? It turns out there are five different sets of information, all of which are sufficient to demonstrate congruence.


Side-side-side congruence (SSS)

If we know that each side of one triangle has a matching congruent side in the other triangle, then the triangles must be congruent. You can try this yourself with three straight objects; you can only make one triangle without changing the lengths, remembering that we are counting mirror images as being congruent as well:

This kind of congruence is called side-side-side, or just SSS.

 


Side-angle-side congruence (SAS)

If we know that two triangles have a pair of matching sides, and the angles between each pair are congruent, then the triangles must be congruent. You can try this yourself with any two straight objects - if you hold them together at a point and a certain angle apart along the rest of their lengths, there is only one triangle you can form by joining the ends together:

This kind of congruence is called side-angle-side, or just SAS.

 

Warning: there is no SSA!

It is possible to have two triangles with a matching pair of sides and a matching angle that are not congruent, like these two:

Using two congruent sides and a congruent angle we have built two obviously non-congruent triangles!

Try using this applet to find the two different triangles with two congruent sides and a congruent angle, just like the picture above:


Hypotenuse-leg congruence (HL)

This congruence test also uses two sides, and is an exception to the "the congruent angle must be between two congruent sides" rule. However, it only applies to right triangles. If two right triangles have congruent hypotenuses and one pair of legs are congruent, then the triangles must be congruent overall.

This congruence test is called hypotenuse-leg, or just HL.


Angle-side-angle and angle-angle-side congruences (ASA and AAS)

What if we only know that one side is congruent? In this case we need to know that there are two pairs of congruent angles. There are different cases that we need to treat separately, the first being when the two given angles lie on either end of the given side:

The dashed line meets the dotted line at exactly one point, so we can only build one triangle from this information.

This kind of congruence is called angle-side-angle congruence, or just ASA.

The other way to use a single pair of congruent sides side to prove congruence of triangles is to know one pair of angles on the congruent sides are equal, and that the angles opposite the congruent sides are also equal:

The dashed line is placed at the correct angle to the segment. If the angle opposite is known, only one of the dotted lines from the other end of the segment will make this particular angle with the dashed line. 

This kind of congruence is called angle-angle-side congruence, or just AAS.

Here is a summary of the five triangle congruence tests.

Proving congruence in triangles

To show that two triangles are congruent, it is sufficient to demonstrate the following:

  • Side-side-side, or SSS
    • The two triangles have three pairs of congruent sides
  • Side-angle-side, or SAS
    • The two triangles have two pairs of congruent sides, and the angles between these sides are also congruent
  • Hypotenuse-leg, or HL
    • The two triangles are right-angled, have congruent hypotenuses, and have one pair of congruent legs
  • Angle-side-angle, or ASA
    • The two triangles have one pair of congruent sides, and two pairs of congruent angles at either end of the congruent sides.
  • Angle-angle-side, or AAS
    • The two triangles have one pair of congruent sides, one pair of congruent angles at the end of the congruent sides, and one pair of congruent angles opposite the congruent sides.

 

Practice questions

QUESTION 4

Consider the following three triangles:

  1. Which of the following triangles are congruent?

    A

    B

    C
  2. State the reason why the two previous triangles are congruent:

    HL: Two right triangles with hypotenuse and one leg are congruent.

    A

    AAS: A pair of corresponding angles and a non-included side are congruent.

    B

    SSS: All three corresponding sides are congruent.

    C

    SAS: A pair of corresponding sides and the included angle are congruent.

    D

QUESTION 5

Consider the given triangles.

Two right triangles are connected at point $C$C, $\triangle ABC$ABC and $\triangle DFC$DFC. They share a common vertex at $C$C but not a common angle such the angle at vertex $C$C of $\triangle ABC$ABC is $\angle BCA$BCA, and the angle at vertex $C$C of $\triangle DFC$DFC is $\angle DCF$DCF.  The triangle on the left is $\triangle ABC$ABC with a right angle, as indicated by a small square at vertex $A$A, between its horizontal base $AC$AC and vertical leg $BA$BA. $\triangle ABC$ABC has angles $\angle ABC$ABC at vertex B and $\angle BCA$BCA at vertex C, and sides $AB$AB and $BC$BC that are not labeled.  The triangle on the right is $\triangle DFC$DFC with a right angle, as indicated by a small square at vertex $D$D, between its horizontal base $DC$DC and vertical leg $FD$FD. Each of the bases of the triangles $AC$AC and $DC$DC have a single tick mark indicating congruence. $\triangle DFC$DFC has angles $\angle DFC$DFC at vertex F and $\angle DCF$DCF at vertex C, and sides $DF$DF and $FC$FC that are not labeled. The right angles at vertex $A$A and vertex $D$D are congruent.
  1. Do we have enough information to deduce that the two triangles are congruent?

    Yes

    A

    No

    B

QUESTION 6

Consider the adjacent figure:

A parallelogram $SRQP$SRQP with vertices $S$S,$R$R, $Q$Q, and $P$P in a clockwise direction starting from $S$S at the upper left vertex. A diagonal is drawn from vertex $S$S to vertex $Q$Q, dividing the parallelogram into two triangles thus also dividing the angles at vertex $S$S and $Q$Q. $\angle QSP$QSP and $\angle SQR$SQR are marked with a double-arc suggesting congruence. $\angle SQP$SQP and $\angle QSR$QSR are marked with a single-arc suggesting congruence.


 
  1. From the information given on the diagram, which angle is congruent to $\angle PSQ$PSQ?

    $\angle PQS$PQS

    A

    $\angle RQS$RQS

    B

    $\angle RSQ$RSQ

    C
  2. State the most direct reason why $\triangle PSQ$PSQ is congruent to $\triangle RQS$RQS.

    SAS: A pair of corresponding sides and the included angle are congruent.

    A

    HL: Two right triangles with hypotenuse and one leg are congruent.

    B

    ASA: A pair of corresponding angles and the included side are congruent.

    C

    AAS: A pair of corresponding angles and a non-included side are congruent.

    D

    SSS: All three corresponding sides are congruent.

    E

Outcomes

I.G.CO.6

Use geometric descriptions of rigid motions to transform figures and to predict the effect of a given rigid motion on a given figure; given two figures, use the definition of congruence in terms of rigid motions to decide if they are congruent.

I.G.CO.7

Use the definition of congruence in terms of rigid motions to show that two triangles are congruent if and only if corresponding pairs of sides and corresponding pairs of angles are congruent.

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