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# Congruent Triangles

# SSS and SAS

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).

In order to see this, we are going to start by fixing three lengths and see that there 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 are congruent because they are 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.

### Sample Problem

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.

Statements | Reasons |

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 ∆*DEF* such that two sides of ∆*ABC* are congruent to two sides of ∆*DEF*. Let's also suppose that the angles between these sides are congruent to one another. For example, suppose *AB* ≅ *DE*, *AC* ≅ *DF*, and ∠*BAC* ≅ ∠*EDF*. If that's the case, then ∆*ABC* ≅ ∆*DEF*. True story.

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.

### Sample Problem

Given: *AB* ≅ *CD*, *AB* || *CD*.

Prove: ∆*ACD* ≅ ∆*ABD*.

Statements | Reasons |

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.

### Sample Problem

Which two triangles, if any, are congruent? How do you know?

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.