# At a Glance - Isosceles Triangle Theorem and Hypotenuse-Leg Theorem

### Sample Problem

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.

Statements | Reasons |

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*.

### Sample Problem

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 side 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.