### Triangle Proportionality Theorem

Think about a **midsegment** of a triangle. A midsegment is parallel to one side of a triangle and divides the other two sides into congruent halves. The midsegment divides those two sides **proportionally.** But what about another line that is parallel, but does not divide the other two sides into congruent halves? In fact, such a line will still divide the sides proportionally. This is called the **Triangle Proportionality Theorem.**

**Triangle Proportionality Theorem:** If a line parallel to one side of a triangle intersects the other two sides, then it divides those sides proportionally.

If \begin{align*}\overline{DE} \| \overline{AC}\end{align*}, then \begin{align*}\frac{BD}{DA} = \frac{BE}{EC}\end{align*}. (\begin{align*}\frac{DA}{BD} = \frac{EC}{BE}\end{align*} *is also a true proportion.*)

The converse of this theorem is also true.

**Triangle Proportionality Theorem Converse:** If a line divides two sides of a triangle proportionally, then it is parallel to the third side.

If \begin{align*}\frac{BD}{DA} = \frac{BE}{EC}\end{align*}, then \begin{align*}\overline {DE} \| \overline{AC}\end{align*}.

What if you were given a triangle with a line segment drawn through it from one side to the other? How could you use information about the triangle's side lengths to determine if that line segment is parallel to the third side?

### Examples

Use the diagram to answers Examples 1 and 2. \begin{align*}\overline{DB} \| \overline{FE}\end{align*}.

#### Example 1

Name the similar triangles. Write the similarity statement.

\begin{align*}\triangle DBC \sim \triangle FEC\end{align*}

#### Example 2

\begin{align*}\frac{FC+?}{FC} = \frac{?}{FE}\end{align*}

DF; DB

#### Example 3

A triangle with its midsegment is drawn below. What is the ratio that the midsegment divides the sides into?

The midsegment splits the sides evenly. The ratio would be 8:8 or 10:10, which both reduce to 1:1.

#### Example 4

In the diagram below, \begin{align*}\overline {EB} \| \overline {CD}\end{align*}. Find \begin{align*}BC\end{align*}.

To solve, set up a proportion.

\begin{align*}\frac{10}{15} = \frac{BC}{12} \longrightarrow \ 15(BC) &= 120\\ BC &= 8\end{align*}

#### Example 5

Is \begin{align*}\overline{DE} \| \overline{CB}\end{align*}?

If the ratios are equal, then the lines are parallel.

\begin{align*}\frac{6}{18} = \frac{8}{24} = \frac{1}{3}\end{align*}

Because the ratios are equal, \begin{align*}\overline {DE} \| \overline{CB}\end{align*}.

### Review

Use the diagram to answer questions 1-7. \begin{align*}\overline{AB} \| \overline {DE}\end{align*}.

- Find \begin{align*}BD\end{align*}.
- Find \begin{align*}DC\end{align*}.
- Find \begin{align*}DE\end{align*}.
- Find \begin{align*}AC\end{align*}.
- What is \begin{align*}BD:DC\end{align*}?
- What is \begin{align*}DC:BC\end{align*}?
- Why \begin{align*}BD:DC \neq DC:BC\end{align*}?

Use the given lengths to determine if \begin{align*}\overline{AB} \| \overline{DE}\end{align*}.

### Review (Answers)

To see the Review answers, open this PDF file and look for section 7.8.

### Resources