Now that we know the criteria for identifying congruent triangles, the next step is to find a way to communicate that knowledge in the form of a proof. Recall two commonly used proof forms, the two-column proof and the paragraph proof.
A two-column proof might be considered the most structured form, where every line of work is placed on a row with an accompanying reason. A generic shell for the proof looks like this:
To prove: $\left[\text{Statement to be proved}\right]$[Statement to be proved] | |
Statement | Reason |
Statement 1 | Reason 1 |
Statement 2 | Reason 2 |
Statement 3 | Reason 3 |
$\ldots$… | $\ldots$… |
$\left[\text{Statement to be proved}\right]$[Statement to be proved] | Reason $n$n |
We use as many lines of reasoning as we need, and make sure the last line always has the statement we want to prove at the end. Any information provided to us should have the reason Given written next to it, to indicate that no reasoning actually took place to arrive at the statement. Let's try a few examples where we prove triangles congruent using a two-column proof.
Show the following two triangles are congruent using a two-column proof:
Think: Before writing anything down, it's a good idea for us to think for a minute about which congruence test we will use. Since none of the angles have been marked, we will have to rely on the sides. The next step (even if you haven't decided on the test yet!) will be to translate the information provided to us into the start of a two-column proof.
Do: We write down what we want to prove, followed by the information obtained from the markings on the diagram:
To prove: $\Delta ADC\cong\Delta CBA$ΔADC≅ΔCBA | |
Statement | Reason |
$\overline{AD}\cong\overline{CB}$AD≅CB | Given |
$\overline{DC}\cong\overline{BA}$DC≅BA | Given |
We have two sides, and by the SSS rule, we only need to show one more pair of sides (one from each triangle) are congruent. In this case the two triangles share a side in common, so we use the reflexive property of congruence to note this:
To prove: $\Delta ADC\cong\Delta CBA$ΔADC≅ΔCBA | |
Statement | Reason |
$\overline{AD}\cong\overline{CB}$AD≅CB | Given |
$\overline{DC}\cong\overline{BA}$DC≅BA | Given |
$\overline{AC}\cong\overline{AC}$AC≅AC | Reflexive property of congruence |
Now we have demonstrated that three sides are the same, we can write our conclusion at the end:
To prove: $\Delta ADC\cong\Delta CBA$ΔADC≅ΔCBA | |
Statement | Reason |
$\overline{AD}\cong\overline{CB}$AD≅CB | Given |
$\overline{DC}\cong\overline{BA}$DC≅BA | Given |
$\overline{AC}\cong\overline{AC}$AC≅AC | Reflexive property of congruence |
$\Delta ADC\cong\Delta CBA$ΔADC≅ΔCBA | Side-side-side congruency (SSS) |
Show the following two triangles are congruent using a two-column proof:
Think: Make sure to take note of every marking on the diagram, and think about what rule to use.
Do: Once again we translate the information provided to us, both what we want to show and what the markings tell us:
To prove: $\Delta AMB\cong\Delta CMD$ΔAMB≅ΔCMD | |
Statement | Reason |
$\overline{AM}\cong\overline{CM}$AM≅CM | Given |
$\overline{BM}\cong\overline{DM}$BM≅DM | Given |
We don't know anything about the third pair of sides, so we will have to use an angle instead. To use the SAS rule, we need to show that the angles lying between the two pairs of congruent sides are also congruent. Here we can use the fact that the angles are vertical, and reach our conclusion:
To prove: $\Delta AMB\cong\Delta CMD$ΔAMB≅ΔCMD | |
Statement | Reason |
$\overline{AM}\cong\overline{CM}$AM≅CM | Given |
$\overline{BM}\cong\overline{DM}$BM≅DM | Given |
$\angle AMB\cong\angle CMD$∠AMB≅∠CMD | Vertical angles |
$\Delta AMB\cong\Delta CMD$ΔAMB≅ΔCMD | Side-angle-side congruency (SAS) |
Show the following two triangles are congruent using a two-column proof:
Think: As usual we make sure to take note of every marking on the diagram, and think about what test would be best to use. Since we are only shown information about a single pair of sides, we need to concentrate on finding two pairs of congruent angles.
Do: Translating what we want to prove and each of the markings in the diagram into the two-column format gives us the following:
To prove: $\Delta PXQ\cong\Delta RXS$ΔPXQ≅ΔRXS | |
Statement | Reason |
$\overline{PQ}\cong\overline{RS}$PQ≅RS | Given |
$\overleftrightarrow{PQ}\parallel\overleftrightarrow{SR}$›‹PQ∥›‹SR | Given |
Having established that $\overleftrightarrow{PQ}$›‹PQ and $\overleftrightarrow{SR}$›‹SR are parallel, we can use the alternate interior angles theorem to show that the angles lying on either end of $\overline{PQ}$PQ and $\overline{RS}$RS are congruent, and reach our conclusion:
To prove: $\Delta PXQ\cong\Delta RXS$ΔPXQ≅ΔRXS | |
Statement | Reason |
$\overline{PQ}\cong\overline{RS}$PQ≅RS | Given |
$\overleftrightarrow{PQ}\parallel\overleftrightarrow{SR}$›‹PQ∥›‹SR | Given |
$\angle XPQ\cong\angle XRS$∠XPQ≅∠XRS | Alternate interior angles theorem |
$\angle XQP\cong\angle XSR$∠XQP≅∠XSR | Alternate interior angles theorem |
$\Delta PXQ\cong\Delta RXS$ΔPXQ≅ΔRXS | Angle-side-angle congruency (ASA) |
The paragraph proof describes the same information, and goes through the same process, as a two-column proof, but we write it all out in more natural language and style. Here are some examples.
Show the following two triangles are congruent using a paragraph proof:
Think: Just like in a two-column proof we want to make sure we know what we want to prove, know what each marking means, and have an idea for a congruence test we want to use in the proof.
Do: We want to prove that $\Delta XAY\cong\Delta XBY$ΔXAY≅ΔXBY. First we can see that $\angle XAY$∠XAY and $\angle XBY$∠XBY are both given as right-angles, so both $\Delta XAY$ΔXAY and $\Delta XBY$ΔXBY are right triangles. We can also see that $\overline{XY}$XY is the hypotenuse of both triangles, so by the reflexive property of congruence, the hypotenuses of both triangles are congruent. Finally, the segments $\overline{XA}$XA and $\overline{XB}$XB are congruent legs, as they are both radii of the circle centered at $X$X. We have therefore shown by Hypotenuse-leg congruency (HL) that $\Delta XAY\cong\Delta XBY$ΔXAY≅ΔXBY, QED.
You may have seen proofs conclude with the letters QED, an abbreviation for the Latin phrase "quod erat demonstrandum". This roughly translates to "I have proved what I set out to prove", but it really means "I delivered on my promise and no-one can argue with me".
It is commonly mispronounced as "quod erat demonstraTum", and modern mathematicians usually use the symbol in its place. Still, it is a powerful phrase and powerful idea, which is why it shows up outside of mathematics - in other academic learning areas, as well as books, films, and other popular media.
Show the following two triangles are congruent using a paragraph proof:
Think: If you're more comfortable with two-column proofs, you might want to write out a list of statements first in that style before writing it in full sentences.
Do: We wish to demonstrate that $\Delta PAC\cong\Delta QBD$ΔPAC≅ΔQBD. We are given that $\overline{PA}$PA and $\overline{QB}$QB are parallel, that $\angle APC$∠APC and $\angle BQD$∠BQD are congruent, and that the segments $\overline{AB}$AB, $\overline{BC}$BC, and $\overline{CD}$CD are all congruent. Since congruent segments have equal measure, we know that $AB=CD$AB=CD, and by the addition property of equality we can conclude that $AB+BC=BC+CD$AB+BC=BC+CD. This then means that the segments $\overline{AC}$AC and $\overline{BD}$BD have equal measure, so they must be congruent. The angles $\angle PAC$∠PAC and $\angle QBD$∠QBD are congruent since they are corresponding angles in the parallel lines $\overleftrightarrow{PA}$›‹PA and $\overleftrightarrow{QB}$›‹QB. We have therefore shown using angle-angle-side congruency (AAS) that $\Delta PAC$ΔPAC and $\Delta QBD$ΔQBD are congruent, QED.
No matter which method of proof you use, showing that two triangles are congruent involves preparing carefully justified statements that line up with one of the tests for congruency. Make sure you know them all and are ready to use them before trying to learn these two new proof methods.
This two-column proof shows that $\Delta ABC\cong\Delta XYZ$ΔABC≅ΔXYZ in the attached diagram, but it is incomplete.
Statements | Reasons |
---|---|
$\overline{AC}\cong\overline{XZ}$AC≅XZ | Given |
$\angle CBA\cong\angle ZYX$∠CBA≅∠ZYX |
Given |
$\angle CAB\cong\angle ZXY$∠CAB≅∠ZXY |
Given |
$\Delta ABC\cong\Delta XYZ$ΔABC≅ΔXYZ |
$\left[\text{____}\right]$[____] |
Select the correct reason to complete the proof.
Angle-angle-side congruence (AAS)
Side-side-side congruence (SSS)
Side-angle-side congruence (SAS)
Side-side-angle congruence (SSA)
Angle-side-angle congruence (ASA)
Angle-angle-side congruence (AAS)
Side-side-side congruence (SSS)
Side-angle-side congruence (SAS)
Side-side-angle congruence (SSA)
Angle-side-angle congruence (ASA)
This two-column proof shows that $\Delta PQR\cong\Delta RSP$ΔPQR≅ΔRSP in the attached diagram, but it is incomplete.
To prove: $\Delta PQR\cong\Delta RSP$ΔPQR≅ΔRSP | |
---|---|
Statements | Reasons |
$\left[\text{__________}\right]$[__________] | $\left[\text{__________}\right]$[__________] |
$\overline{PR}\cong\overline{RP}$PR≅RP | Reflexive property of congruence |
$\angle QRP\cong\angle SPR$∠QRP≅∠SPR |
Given |
$\Delta PQR\cong\Delta RSP$ΔPQR≅ΔRSP |
Side-angle-side congruence (SAS) |
Select the correct reason to complete the proof.
$\overline{PS}\cong\overline{RQ}$PS≅RQ | Transitive property of congruence |
$\overline{SR}\cong\overline{QP}$SR≅QP | Given |
$\overline{PS}\cong\overline{RQ}$PS≅RQ | Given |
$\overline{SR}\cong\overline{QP}$SR≅QP | Transitive property of congruence |
$\overline{PS}\cong\overline{RQ}$PS≅RQ | Transitive property of congruence |
$\overline{SR}\cong\overline{QP}$SR≅QP | Given |
$\overline{PS}\cong\overline{RQ}$PS≅RQ | Given |
$\overline{SR}\cong\overline{QP}$SR≅QP | Transitive property of congruence |
This two-column proof shows that $\Delta DEH\cong\Delta FEG$ΔDEH≅ΔFEG 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 |
$\left[\text{___}\right]$[___] |
Select the correct pair of reasons to complete the proof.
$\angle EDH\cong\angle FEG$∠EDH≅∠FEG | Corresponding angles theorem |
and | |
$\Delta DEH\cong\Delta FEG$ΔDEH≅ΔFEG | Side-side-angle congruence (SSA) |
$\overline{DE}\cong\overline{EF}$DE≅EF | Reflexive property of congruence |
and | |
$\Delta DEH\cong\Delta FEG$ΔDEH≅ΔFEG | Side-side-side congruence (SSS) |
$\angle DHE\cong\angle FGE$∠DHE≅∠FGE | Triangle hinge theorem |
and | |
$\Delta DEH\cong\Delta FEG$ΔDEH≅ΔFEG | Side-angle-side congruence (SAS) |
$\overline{DE}\cong\overline{EF}$DE≅EF | Properties of a midpoint |
and | |
$\Delta DEH\cong\Delta FEG$ΔDEH≅ΔFEG | Side-side-side congruence theorem (SSS) |
$\angle EDH\cong\angle FEG$∠EDH≅∠FEG | Corresponding angles theorem |
and | |
$\Delta DEH\cong\Delta FEG$ΔDEH≅ΔFEG | Side-side-angle congruence (SSA) |
$\overline{DE}\cong\overline{EF}$DE≅EF | Reflexive property of congruence |
and | |
$\Delta DEH\cong\Delta FEG$ΔDEH≅ΔFEG | Side-side-side congruence (SSS) |
$\angle DHE\cong\angle FGE$∠DHE≅∠FGE | Triangle hinge theorem |
and | |
$\Delta DEH\cong\Delta FEG$ΔDEH≅ΔFEG | Side-angle-side congruence (SAS) |
$\overline{DE}\cong\overline{EF}$DE≅EF | Properties of a midpoint |
and | |
$\Delta DEH\cong\Delta FEG$ΔDEH≅ΔFEG | Side-side-side congruence theorem (SSS) |
Use congruence and similarity criteria for triangles to solve problems algebraically and geometrically.
Use congruence and similarity criteria for triangles to prove relationships in geometric figures.