What if two mountain bikers leave from the same parking lot and head in opposite directions on two different trails? The first rider goes 8 miles due west, then rides due south for 15 miles. The second rider goes 6 miles due east, then changes direction and rides \begin{align*}20^\circ\end{align*} east of due north for 17 miles. Both riders have been traveling for 23 miles, but which one is further from the parking lot?

### Comparing Angles and Sides in Triangles

Look at the triangle below. The sides of the triangle are given. Can you determine which angle is the largest? As you might guess, the largest angle will be opposite 18 because it is the longest side. Similarly, the smallest angle will be opposite the shortest side, 7. Therefore, the angle measure in the middle will be opposite 13.

**Theorem:** If one side of a triangle is longer than another side, then the angle opposite the longer side will be larger than the angle opposite the shorter side.

**Converse:** If one angle in a triangle is larger than another angle in a triangle, then the side opposite the larger angle will be longer than the side opposite the smaller angle.

*Proof of Theorem:*

Given: \begin{align*}AC > AB\end{align*}

Prove: \begin{align*}m \angle ABC > m \angle C\end{align*}

Statement |
Reason |
---|---|

1. \begin{align*}AC > AB\end{align*} | Given |

2. Locate point \begin{align*}P\end{align*} such that \begin{align*}AB = AP\end{align*} | Ruler Postulate |

3. \begin{align*}\triangle ABP\end{align*} is an isosceles triangle | Definition of an isosceles triangle |

4. \begin{align*}m \angle 1 = m \angle 3\end{align*} | Base Angles Theorem |

5. \begin{align*}m \angle 3 = m \angle 2 + m \angle C\end{align*} | Exterior Angle Theorem |

6. \begin{align*}m \angle 1 = m \angle 2 + m \angle C\end{align*} | Substitution PoE |

7. \begin{align*}m \angle ABC = m \angle 1 + m \angle 2\end{align*} | Angle Addition Postulate |

8. \begin{align*}m \angle ABC = m \angle 2 + m \angle 2 + m \angle C\end{align*} | Substitution PoE |

9. \begin{align*}m \angle ABC > m \angle C\end{align*} | Definition of “greater than” (from step 8) |

We have two congruent triangles \begin{align*}\triangle ABC\end{align*} and \begin{align*}\triangle DEF\end{align*}, marked below:

Therefore, if \begin{align*}AB = DE\end{align*} and \begin{align*}BC = EF\end{align*} and \begin{align*}m \angle B >m \angle E\end{align*}, then \begin{align*}AC > DF\end{align*}. Now, let’s adjust \begin{align*}m \angle B >m \angle E\end{align*}. Would that make \begin{align*}AC > DF\end{align*}? Yes. See the picture below.

**The SAS Inequality Theorem (Hinge Theorem):** If two sides of a triangle are congruent to two sides of another triangle, but the included angle of one triangle has greater measure than the included angle of the other triangle, then the third side of the first triangle is longer than the third side of the second triangle.

**SSS Inequality Theorem (also called the Converse of the Hinge Theorem):** If two sides of a triangle are congruent to two sides of another triangle, but the third side of the first triangle is longer than the third side of the second triangle, then the included angle of the first triangle is greater in measure than the included angle of the second triangle.

#### Categorizing Length

List the sides in order, from shortest to longest.

First, we need to find \begin{align*}m \angle A\end{align*}. From the Triangle Sum Theorem, \begin{align*}m \angle A + 86^\circ + 27^\circ = 180^\circ\end{align*}. So, \begin{align*}m \angle A = 67^\circ\end{align*}. Therefore, we can conclude that the longest side is opposite the largest angle. \begin{align*}86^\circ\end{align*} is the largest angle, so \begin{align*}AC\end{align*} is the longest side. The next largest angle is \begin{align*}67^\circ\end{align*}, so \begin{align*}BC\end{align*} would be the next longest side. \begin{align*}27^\circ\end{align*} is the smallest angle, so \begin{align*}AB\end{align*} is the shortest side. In order from shortest to longest, the answer is: \begin{align*}AB, BC, AC\end{align*}.

#### Listing Angles in Order

List the angles in order, from largest to smallest.

Just like with the sides, the largest angle is opposite the longest side. The longest side is \begin{align*}BC\end{align*}, so the largest angle is \begin{align*}\angle A\end{align*}. Next would be \begin{align*}\angle B\end{align*} and finally \begin{align*}\angle A\end{align*} is the smallest angle.

#### Listing Sides in Order

List the sides in order, from least to greatest.

Let’s start with \begin{align*}\triangle DCE\end{align*}. The missing angle is \begin{align*}55^\circ\end{align*}. Therefore the sides, in order are \begin{align*}CE, CD\end{align*}, and \begin{align*}DE\end{align*}. For \begin{align*}\triangle BCD\end{align*}, the missing angle is \begin{align*}43^\circ\end{align*}. The order of the sides is \begin{align*}BD, CD\end{align*}, and \begin{align*}BC\end{align*}. By the SAS Inequality Theorem, we know that \begin{align*}BC > DE\end{align*}, so the order of all the sides would be: \begin{align*}BD = CE, CD, DE, BC\end{align*}.

#### Parking Lot Problem Revisited

Even though the two sets of lengths are not equal, they both add up to 23. Therefore, the second rider is further away from the parking lot because \begin{align*}110^\circ> 90^\circ\end{align*}.

### Examples

#### Example 1

If \begin{align*}\overline{XM}\end{align*} is a median of \begin{align*}\triangle XYZ\end{align*} and \begin{align*}XY > XZ\end{align*}, what can we say about \begin{align*}m \angle 1\end{align*} and \begin{align*}m \angle 2\end{align*}?

By the definition of a median, \begin{align*}M\end{align*} is the midpoint of \begin{align*}\overline{YZ}\end{align*}. This means that \begin{align*}YM = MZ\end{align*}. \begin{align*}MX = MX\end{align*} by the Reflexive Property and we know that \begin{align*}XY > XZ\end{align*}. Therefore, we can use the SSS Inequality Theorem to conclude that \begin{align*}m \angle 1 >m \angle 2\end{align*}.

#### Example 2

List the sides of the two triangles in order, from least to greatest.

Here we have no congruent sides or angles. So, let’s look at each triangle separately. Start with \begin{align*}\triangle XYZ\end{align*}. First the missing angle is \begin{align*}42^\circ\end{align*}. The order of the sides is \begin{align*}YZ, XY\end{align*}, and \begin{align*}XZ\end{align*}. For \begin{align*}\triangle WXZ\end{align*}, the missing angle is \begin{align*}55^\circ\end{align*}. The order of these sides is \begin{align*}XZ, WZ\end{align*}, and \begin{align*}WX\end{align*}. Because the longest side in \begin{align*}\triangle XYZ\end{align*} is the shortest side in \begin{align*}\triangle WXZ\end{align*}, we can put all the sides together in one list: \begin{align*}YZ, XY, XZ, WZ, WX\end{align*}.

#### Example 3

Below is isosceles triangle \begin{align*}\triangle ABC\end{align*}. List everything you can about the triangle and why.

\begin{align*}AB = BC\end{align*} because it is given, \begin{align*}m \angle A = m \angle C\end{align*} by the Base Angle Theorem, and \begin{align*}AD < DC\end{align*} because \begin{align*}m \angle ABD < m \angle CBD\end{align*} and the SAS Triangle Inequality Theorem.

### Interactive Practice

### Review

For questions 1-3, list the sides in order from shortest to longest.

For questions 4-6, list the angles from largest to smallest.

- Compare \begin{align*}m \angle 1\end{align*} and \begin{align*}m \angle 2\end{align*}.
- List the sides from shortest to longest.
- Compare \begin{align*}m \angle 1\end{align*} and \begin{align*}m \angle 2\end{align*}. What can you say about \begin{align*}m \angle 3\end{align*} and \begin{align*}m \angle 4\end{align*}?

In questions 10-12, compare the measures of \begin{align*}a\end{align*} and \begin{align*}b\end{align*}.

In questions 13 and 14, list the measures of the sides in order from least to greatest

In questions 15 and 16 is the conclusion true or false?

- Conclusion: \begin{align*}m \angle C < m \angle B < m \angle A\end{align*}
- Conclusion: \begin{align*}AB < DC\end{align*}
- If \begin{align*}\overline{AB}\end{align*} is a median of \begin{align*}\triangle CAT\end{align*} and \begin{align*}CA>AT\end{align*}, explain why \begin{align*}\angle ABT\end{align*} is acute. You may wish to draw a diagram.

### Review (Answers)

To view the Review answers, open this PDF file and look for section 5.6.