Geometry (A): READ: Midsegment Reading

Midsegments of a triangle

Learning Objectives

Identify the midsegment of a triangle.
Apply the Midsegment Theorem to solve problems involving side lengths and midsegments of triangles.
Use the Midsegment Theorem to solve problems involving variable side lengths and midsegments of triangles.

Introduction

In previous lessons, we used the parallel postulate to learn new theorems that enabled us to solve a variety of problems about parallel lines:

Parallel Postulate: Given: line $l$ and a point $P$ not on $l$ . There is exactly one line through $P$ that is parallel to $l$ .

In this lesson we extend these results to learn about special line segments within triangles. For example, the following triangle contains such a configuration:

Triangle $\triangle XYZ$ is cut by $\overline{AB}$ where $A$ and $B$ are midpoints of sides $\overline{XZ}$ and $\overline{YZ}$ respectively. $\overline{AB}$ is called a midsegment of $\triangle XYZ$ . Note that $\triangle XYZ$ has other midsegments in addition to $\overline{AB}$ . Can you see where they are in the figure above?

If we construct the midpoint of side $\overline{XY}$ at point $C$ and construct $\overline{CA}$ and $\overline{CB}$ respectively, we have the following figure and see that segments $\overline{CA}$ and $\overline{CB}$ are midsegments of $\triangle XYZ$ .

In this lesson we will investigate properties of these segments and solve a variety of problems.

Properties of midsegments within triangles

We start with a theorem that we will use to solve problems that involve midsegments of triangles.

Midsegment Theorem: The segment that joins the midpoints of a pair of sides of a triangle is:

parallel to the third side.
half as long as the third side.

Proof of 1. We need to show that a midsegment is parallel to the third side. We will do this using the Parallel Postulate.

Consider the following triangle $\triangle XYZ$ . Construct the midpoint $A$ of side $\overline{XZ}$ .

By the Parallel Postulate, there is exactly one line though $A$ that is parallel to side $\overline{XY}$ . Let’s say that it intersects side $\overline{YZ}$ at point $B$ . We will show that $B$ must be the midpoint of $\overline{XY}$ and then we can conclude that $\overline{AB}$ is a midsegment of the triangle and is parallel to $\overline{XY}$ .

We must show that the line through $A$ and parallel to side $\overline{XY}$ will intersect side $\overline{YZ}$ at its midpoint. If a parallel line cuts off congruent segments on one transversal, then it cuts off congruent segments on every transversal. This ensures that point $B$ is the midpoint of side $\overline{YZ}$ .

Since $\overline{XA} \cong \overline{AZ}$ , we have $\overline{BZ} \cong \overline{BY}$ . Hence, by the definition of midpoint, point $B$ is the midpoint of side $\overline{YZ}$ . $\overline{AB}$ is a midsegment of the triangle and is also parallel to $\overline{XY}$ .

Proof of 2. We must show that $AB = \frac{1} {2} XY$ .

In $\triangle XYZ$ , construct the midpoint of side $\overline{XY}$ at point $C$ and midsegments $\overline{CA}$ and $\overline{CB}$ as follows:

First note that $\overline{CB} \parallel \overline{XZ}$ by part one of the theorem. Since $\overline{CB}\parallel\overline{XZ}$ and $\overline{AB} \parallel\overline{XY}$ , then $\angle{XAC} \cong \angle{BCA}$ and $\angle{CAB} \cong \angle{ACX}$ since alternate interior angles are congruent. In addition, $\overline{AC} \cong \overline{CA}$ .

Hence, $\triangle{AXC} \cong \triangle{CBA}$ by The ASA Congruence Postulate. $\overline{AB} \cong \overline{XC}$ since corresponding parts of congruent triangles are congruent. Since $C$ is the midpoint of $\overline{XY}$ , we have $XC = CY$ and $XY = XC + CY = XC + XC = 2AB$ by segment addition and substitution.

So, $2AB = XY$ and $AB = \frac{1} {2} XY$ . $\blacklozenge$

Example 1

Use the Midsegment Theorem to solve for the lengths of the midsegments given in the following figure.

$M$ , $N$ and $O$ are midpoints of the sides of the triangle with lengths as indicated. Use the Midsegment Theorem to find

: B. The perimeter of the triangle $\triangle XYZ$ .
: A. $MN$ .

: B. By the Midsegment Theorem, $OM = 3$ implies that $ZY = 6$ ; similarly, $XZ = 8$ , and $XY = 10$ . Hence, the perimeter is $6 + 8 + 10 = 24.$
: A. Since $O$ is a midpoint, we have $XO = 5$ and $XY = 10$ . By the theorem, we must have $MN = 5$ .

We can also examine triangles where one or more of the sides are unknown.

Example 2

Use the Midsegment Theorem to find the value of $x$ in the following triangle having lengths as indicated and midsegment $\overline{XY}$ .

By the Midsegment Theorem we have $2x - 6 = \frac{1} {2}(18)$ . Solving for $x$ , we have $x = \frac{15} {2}$ .

Lesson Summary

In this lesson we:

Introduced the definition of the midsegment of a triangle and examined examples.
Stated and proved the Midsegment Theorem.
Solved problems using the Midsegment Theorem.

Last modified: Monday, June 28, 2010, 3:45 PM