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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
Side Splitter Theorem Picture

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?

formula non example

$ \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?

side splitter diagram

$ \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 corollary to side splitter theorem example, picture Example 2 Side Splitter full

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

Find the length of VX by using the side splitter theorem.

Practice Problem side splitter theorem

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

Use the corollary to find the value of x in the problem pictured below.

problem on corollary to side splitter theorem

Set up the proportion then solve for x:

ratio answer
Problem 3

Use the corollary to find the value of x in the problem pictured below.

problem on corollary to side splitter theorem
ratio answer
Problem 4

Are the red segments pictured below parallel? (Picture not to scale).

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.

Back to Similar Triangles Next to Angle Bisector Theorem