# Side Splitter Theorem

A theorem to find sides of similar triangles

#### What is the side splitter theorem?

The side splitter theorem states that if a line is parallel to a side of a triangle and the line intersects the other two sides, then this line divides those two sides proportionally.

The side splitter theorem is a natural extension of similarity ratio, and it happens any time that a pair of parallel lines intersect a triangle.

Diagram 1

The Side Splitter theorem states that when parallel lines (or segments) like $$BC$$ and $$DE$$ intersect sides of overlapping triangles like $$AD$$ and $$AE$$ , then the intercepted segments are proportional .

NOTE: The side splitter only applies to the intercepted sides. It does not apply to the "bottoms".

#### Is the proportion below true?

$\frac{LP}{PO} = \frac{LM}{MN}$

No, this example is not accurate. PM is obviously not parallel to OM.

Therefore, the side splitter theorem does not hold and is not true.

$\frac{LP}{PO} \color{Red}{\ne} \frac{LM}{MN}$

#### Is the proportion below true?

$\frac{VW}{WY} = \frac{WX}{YZ}$

No, remember this theorem only applies to the segments that are 'split' or intercepted by the parallel lines.

$\frac{VW}{WY} \color{Red}{\ne} \frac{WX}{YZ}$

Instead, you could set up the following proportion:

$\frac{VW}{WY} = \frac{VX}{XZ}$

#### What if there are more than two parallel lines?

Answer: A corollary of the this theorem is that when three parallel lines intersect two transversals, then the segments intercepted on the transversal are proportional.

Example 1 Example 2

### Interactive Demonstration

$$\frac{AC}{CE} = \frac{AB}{BD} \\ \frac{\class{side-ac}{3.86}}{\class{side-ce}{3.86}} = \frac{\class{side-ab}{2.35}}{\class{side-bd}{2.35}} \\ \class{side-final}{1} = \class{side-final}{1} \\$$
Drag Circles or BC Line to Start Demonstration

### Practice Problems

##### Problem 1

To solve this problem, set up the following proportion and solve:

$$\frac{VW}{WY} = \frac{VX}{XZ} \\ \frac{7}{14} = \frac{VX}{16} \\ \frac{16 \cdot 7}{14} = VX \\ VX = 8$$
##### Problem 2

Set up the proportion then solve for x:

##### Problem 4

If the red segments are parallel, then they 'split' or divide triangle's sides proportionally. However, when you try to set up the proportion, you will se that it is not true:

$\frac{10}{12} \color{Red}{\ne} \frac{8}{14}$

Therefore, the red segments are not parallel.