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When two triangles are congruent, they're identical in every single way. Sure, they might be flipped or turned on their side or a million miles away, but they're still clones of each other. If two triangles are congruent, then we should be able to perform only congruence transformations in order to map one triangle onto the other.
That's all good and fine, but it doesn't help us when we're given side lengths and angle measures. What helps is knowing that when triangles are congruent, all their angles and sides are congruent too. Because we know that's true, we can say that corresponding parts of congruent triangles are congruent. We shorten this to CPCTC, just for fun. Kind of like the definition of what congruent triangles actually are.
Do we really have to check all of the edges and the angles to prove congruence? That is a lot of stuff to check, and we all have lives. Even the triangles have places to go and people to see. Isn't there a faster way?
We're glad we asked! (We wouldn't have had to ask if you'd just done it.) Yes, there is a better way. Actually, there are four better ways.
If two triangles have edges with the exact same lengths, then these triangles are congruent. This is called the Side Side Side Postulate, or SSS for short (not to be confused with the Selective Service System).
It's like saying that if two Oompa-Loompas wear clothes with all the same measurements, they're identical. It's pretty much true if you think about it, since all the Oompa-Loompas are practically identical anyway.
Although this seems intuitively clear, let's see if we can figure out why this is a reasonable thing to assume. What this postulate is really saying is that a triangle is completely determined by its three sides (not counting reflection or rotations).
Let's start by fixing three lengths and show that there's only one triangle that we can draw whose sides have those three lengths. To start with, we have three distinct lengths a, b, and c.
We first draw a line with the same length as line a. Label this piece BC.
Now, we need to connect another side of length b to one end of BC. Let's choose to connect at point B. There are a lot of ways to orient the second side once it's connected to point B. To visualize all of the ways to do this, we draw a circle centered at B, and with radius length b.
By definition, every point on this circle creates a line of length b when connected to the point B. That means we have a whole lot of different options for the triangle at this point. So far, so good.
Now we draw a circle centered at the point C with radius c. These points all correspond to possible ways of drawing the third side with the length we want.
How many ways can you draw a triangle with side lengths a, b and c? In order for a triangle to have all these side lengths, the third vertex must lie on both circles, to guarantee that the second and third sides are of length b and c. These circles only intersect at two points. Drawing the triangles that correspond to these two intersection points, we see that there are only two triangles possible.
We're sorry to burst your bubble, but the two triangles we drew are really just one. They're congruent because they're mirror images of each other. So actually, they're the same triangle.
There's our evidence that three sides of a triangle completely determine the shape of the triangle itself. That is, just choosing the lengths of the sides predetermines all of the side lengths and angles. Just like Oompa-Loompas are determined, among other things, by the clothes they wear. Seriously, those white overall things? So 1971.
Given: Suppose ∆ABC is isosceles. That is, AB ≅ BC.
Prove: ∠CAB ≅ ∠ACB.
First, we can bisect AC with our straightedge and compass. We'll name the intersection point D. That will give us AD and DC, both of which are congruent to one another. If we draw a segment from B to D, we'll have split ∆ABC into two triangles: ∆ABD and ∆CBD.
Since it's given that AB ≅ BC, we know that BD is congruent to itself, and AD ≅ DC because of the definition of a bisector, we can use the Side Side Side Postulate to say that ∆ABD ≅ ∆CBD. In that case, ∠CAB ≅ ∠ACB because corresponding angles in congruent triangles must be congruent.
|1. Bisect AC with midpoint D||1. Construction|
|2. AD ≅ DC||2. Definition of Bisected Segment|
|3. Draw in BD||3. Construction|
|4. BD ≅ BD||4. Reflexive Property|
|5. AB ≅ BC||5. Given|
|6. ∆ABD ≅ ∆CBD||6. SSS Postulate (2, 4, and 5)|
|7. ∠CAB ≅ ∠ACB||7. Definition of Congruent Triangles|
What we've done is prove not only that the Side Side Side Postulate works, but also that isosceles triangles have two congruent angles as well as two congruent sides. It's like the premiere of a movie sequel. Isosceles II: This Time It's Angular. Probably not Academy Award-winning, but it should prove useful either way.
The other method we can use for proving triangle congruence is the Side Angle Side Postulate. We also call it SAS method. Giving your teachers SAS will get you an A, but giving your teachers "sass" will get you a one-way ticket to the principal's office.
Suppose we have two triangles, ∆ABC and ∆DFE such that two sides of ∆ABC are congruent to two sides of ∆DFE. Let's also suppose that the angles between these sides are congruent to one another. For example, suppose AB ≅ DF, AC ≅ DE, and ∠BAC ≅ ∠EDF. If that's the case, then ∆ABC ≅ ∆DFE. True story. We don't even need to worry about that third side or the other two angles.
We can think of this postulate fixing the lengths of two sides of a triangle and saying that the length of the third side is determined by the angle opposite to it. We'll cover this much more when we get into trigonometry. Let's just hope we'll get out of trigonometry, too.
Given: AB ≅ CD, AB || CD.
Prove: ∆ACD ≅ ∆ABD.
|1. AB ≅ CD||1. Given|
|2. AB || CD||2. Given|
|3. ∠BAD ≅ ∠CDA||3. Alternate Interior Angles Theorem|
|4. AD ≅ AD||4. Reflexivity of Congruence|
|5. ∆ACD ≅ ∆ABD||5. SAS Postulate|
What's important to remember about SAS is that, like the name suggests, the angle we're using must be between the two sides. We call it the included angle. If we use any other angle, we won't be able to prove that the triangles are congruent, which will make us sad.
Which two triangles, if any, are congruent? How do you know?
If we go alphabetically, it'll be easier to keep track of all this crazy triangle business we've got going on. Maybe we should brush up on our ABC's.
∆ABC has two segments and a defined angle in between them. The only other triangle with those same segments is ∆GHI. Luckily, the angle in between those segments is congruent to that of ∆ABC. That means we can say that ∆ABC ≅ ∆GHI by the SAS postulate.
Next on our list is an obtuse triangle, ∆DEF. The only other obtuse triangle with the same angle measure (also the only other obtuse triangle at all) is ∆KLJ. While both their obtuse angles are 140°, the lengths on either side of the angles aren't congruent. That means ∆DEF ≇ ∆KLJ.
What about ∆MNO? All its sides are 13 units in length. The only other triangle with side lengths of 13 units is ∆PQR, but we're only given two side lengths, not three. What ever will we do? Well, ∆MNO is an equilateral triangle (by definition), so all its angles are 60°. Since the angle we have in ∆PQR is 60°, we have enough information using the SAS and/or SSS postulate to say that ∆MNO ≅ ∆PQR.
Our last pair of triangles is ∆STU and ∆WVX. They both have right angles and side lengths that match one another. Since the angles are in between the defined side lengths and they're all congruent, ∆STU ≅ ∆WVX because of the SAS postulate.
Up until now, sides have been hogging the spotlight (they're really hammy). If we give angles the floor for a bit, maybe they'll make some interesting discoveries for us.
If we have two angles whose sum is strictly less than 180° and we draw them on either side of a segment, we'd have something that looks like this.
Imagine pointing a laser off of each end at the desired angle. There's only one point where these laser lines will intersect. In other words, specifying two angles and the length of the line segment between them can give us only one triangle.
That means we can determine whether two triangles are congruent by knowing two angles and the included side. That's called the Angle Side Angle Postulate, abbreviated ASA (or ASAP, which is when we should use it).
Is it true that ∆ABC ≅ ∆DEF? How do you know?
Let's take a look at what we have. We have two sets of congruent angles: ∠A ≅ ∠D and ∠C ≅ ∠F. The segments in between them, AC and DF, are also congruent. If all of that is true (which it is), then we can say that ∆ABC ≅ ∆DEF by ASA (and no, we don't mean the American Society of Anesthesiologists).
One important fact to consider: Knowing two angles in a triangle can automatically give us the third angle, thanks to the triangle Angle Sum Theorem.
This means that knowing any two angles and one side is essentially the same as the ASA postulate. Since the only other arrangement of angles and sides available is two angles and a non-included side, we call that the Angle Angle Side Theorem, or AAS.
A quick thing to note is that AAS is a theorem, not a postulate. Since we use the Angle Sum Theorem to prove it, it's no longer a postulate because it isn't assumed anymore. Basically, the Angle Sum Theorem for triangles elevates its rank from postulate to theorem.
Don't believe us?
In the image, we can see that AC ≅ DF. We also know that ∠CAB ≅ ∠FDE and ∠ABC ≅ ∠DEF. So this is a generic AAS situation.
The triangle Angle Sum Theorem tells us that all the interior angles in a triangle add up to 180°. We can rewrite that into equations.
m∠BCA + m∠ABC + m∠CAB = 180
m∠EFD + m∠DEF + m∠FDE = 180
Solving for ∠BCA and ∠EFD shouldn't be a problem.
m∠BCA = 180 – m∠ABC – m∠CAB
m∠EFD = 180 – m∠DEF – m∠FDE
It follows by substitution that ∠BCA ≅ ∠EFD.
Since knowing any two angles is equivalent to knowing all 3 angles, the ASA postulate can be generalized to this: If we have two triangles ∆ABC and ∆DEF and AB is congruent to DE, the two triangles are congruent as long as any of two corresponding angles are congruent.
Given: ∠BAC ≅ ∠ACB.
Prove: ∆ABC is isosceles because AB ≅ BC.
Remember how we proved that isosceles triangles have two congruent angles because they have two congruent sides? This proof is asking us to do the exact opposite. What do these proofs want from us and why can't they just leave us alone? Buck up, buddy. We promise it won't be that bad.
What we can do is bisect the unknown angle. That'll split ∆ABC into two triangles with BD as a shared side. The angle bisector will make two congruent angles, one in each smaller triangle. The given angles, ∠BAC and ∠ACB, are congruent. The side shared by both triangles is definitely congruent to itself.
So what do we have? Two angles and a side in between them for both triangles—each one congruent to the other triangle's corresponding part. What we have here is the recipe for ASA. Since both the triangles are congruent, all their sides are congruent. Need we say more?
We don't need to, but we'll give you the formal proof just in case.
|1. ∠BAC ≅ ∠ACB||1. Given|
|2. Bisect ∠ABC with BD||2. Construction|
|3. ∠ABD ≅ ∠DBC||3. Definition of Angle Bisector|
|4. BD ≅ BD||4. Reflexive Property|
|5. ∆ABD ≅ ∆CBD||5. AAS (1, 3, and 4)|
|6. AB ≅ BC||6. Definition of Congruence|
|7. ∆ABC is isosceles||7. Definition of Isosceles Triangle|
What we've done is prove something pretty darn important about isosceles triangles. The angles opposite their congruent sides are also congruent. Since we've now proved this idea inside and out, we can finally give it a name: the Isosceles Triangle Theorem. It says: If we're given ∆ABC, AB ≅ BC if and only if ∠BAC ≅ ∠BCA.
Which triangle is congruent to ∆ABC?
If their sides and angles match up, they're congruent. So let's see if their sides and angles match up. Sounds easy enough.
We'll look at ∆DEF first. Two of its angles are congruent to ∆ABC (∠D ≅ ∠A and ∠E ≅ ∠C), and the segment in the middle is also congruent to one of the segments of ∆ABC (DE ≅ BC). For the triangles to be congruent, we'd need the corresponding segments to be congruent, so DE would need to be congruent to AC, not BC. That means they aren't congruent because ASA isn't satisfied.
How about ∆JKL? We can see clearly that ∠A ≅ ∠J and ∠C ≅ ∠L, but JK ≅ AC when according to AAS, it should be congruent to AB. That's no good either.
Our last hope is ∆GHI. We know ∠G ≅ ∠A and ∠H ≅ ∠B. By extension, that means ∠I ≅ ∠C (another shout out to the Angle Sum Theorem). We're given that GI ≅ AC, which is exactly what we need for the triangles to be congruent. Because of AAS, ∆ABC ≅ ∆GHI.
Angles really stepped up to the plate with ASA and AAS, and sides served us well with SSS and SAS. Like a bankrupt farmer, we want to milk them for all they're worth. What if they're holding out on us? What about postulates they haven't yet unveiled, like AAA or SSA?
Well, we may have been suspicious for no reason because those postulates don't exist.
From the figure above, it is pretty clear that you need to know that at least one side length is congruent, or you can have two triangles that are of completely different sizes but they have the same angles. While these triangles are not congruent, later on we'll talk more about what sort of relationship these two have (and maybe even make it Facebook-official).
Well, AAA doesn't work. That doesn't bode well for the insurance company. What about SSA (or its significantly more vulgar alter-ego)? You'd think that SSA would be just as good as any other postulate, but nope. Rather than simply tell you why, we'll show you. (No mooning jokes, please.)
In the figure, ∆ABC and ∆ABD both share ∠DAB and AB. Even though BC is congruent to BD, these triangles are clearly not congruent. We have visual proof, but we can also calculate that AC is not congruent to AD.
On the other hand, SSA does work for a very specific kind of triangle: right triangles. If we know that two triangles are right (have 90° angles) and we know the length of the hypotenuse and one leg on each triangle, this is enough to find the length of the remaining leg using the Pythagorean Theorem.
Rather than calculate the remaining leg and prove triangle congruence using SSS, we'll call this specialized SSA the Hypotenuse-Leg Theorem, implying that it only works if we know the lengths of the hypotenuses and corresponding leg for a pair of right triangles.