USA - Geometry Similarity

Triangle Proportionality theorem

20 practice questions 2 video lessons Theory + worked examples

The Triangle Proportionality Theorem

California Geometry • Standard G-SRT.4 • Similarity

The Triangle Proportionality Theorem is a topic in Similarity in the California Common Core State Standards. It is aligned to Standard G-SRT.4, which requires students to prove theorems about triangles, including a line parallel to one side dividing the other two proportionally.

The triangle proportionality theorem states a line parallel to one side of a triangle divides the other two sides proportionally.

Practice 20 questions

Practice questions

Every question with a fully worked solution.

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Watch 2 video(s)

Triangle Proportionality Theorem Watch
Triangle Proportionality Theorem (With Examples) Watch

Theory

The Triangle Proportionality Theorem (also called the side-splitter theorem) says: a line parallel to one side of a triangle divides the other two sides into proportional segments.

If \(DE\parallel AB\) in \(\triangle ABC\), then \(\dfrac{CD}{DA}=\dfrac{CE}{EB}\).

The converse is also true: if a line divides two sides proportionally, it is parallel to the third side.

It comes from AA similarity: the parallel line creates a smaller triangle similar to the whole.

\(DE\parallel AB\) splits the sides proportionally.

The proportion the theorem gives.

The theorem and its converse:

\[DE\parallel AB \iff \dfrac{CD}{DA}=\dfrac{CE}{EB}\]

a line is parallel to the base exactly when it splits the other two sides in equal ratios

Set up the proportion and cross-multiply to find a missing segment.

How to use the theorem

Confirm the line is parallel to a side.
Match the two pieces of each split side.
Write the proportion \(\dfrac{CD}{DA}=\dfrac{CE}{EB}\).
Cross-multiply and solve.

Example 1 — Set up the proportion

In \(\triangle ABC\), \(DE\parallel AB\). If \(CD=4\), \(DA=6\), and \(CE=6\), find \(EB\).

Solution

The parallel line divides the sides proportionally.

\(\dfrac{CD}{DA}\)	\(=\)	\(\dfrac{CE}{EB}\)
\(\dfrac{4}{6}\)	\(=\)	\(\dfrac{6}{EB}\)
\(4\cdot EB\)	\(=\)	\(36\)
\(EB\)	\(=\)	\(9\)

EB equals 9

Example 2 — Solve for x

A line parallel to a side gives \(\dfrac{x}{8}=\dfrac{6}{12}\). Find \(x\).

Solution

Cross-multiply and solve.

\(12x\)	\(=\)	\(48\)
\(x\)	\(=\)	\(4\)

x equals 4

Example 3 — Is the line parallel?

A segment gives \(\dfrac{CD}{DA}=\dfrac{3}{5}\) and \(\dfrac{CE}{EB}=\dfrac{3}{5}\). Is it parallel to the base?

Solution

By the converse, equal ratios mean the segment is parallel to the third side.

yes, the segment is parallel to the base

Example 4 — Midsegment as a special case

What does the theorem give when the parallel line passes through the midpoints?

Solution

Equal parts on both sides — the midsegment, which is half the base.

it gives the midsegment, half the base

Common pitfalls

Match the pieces correctly: top-to-bottom on both sides, or whole-to-part consistently.

The line must be parallel for the proportion to hold; check first.

Use the converse to prove parallel, the theorem to find lengths.

Frequently asked questions

What is the triangle proportionality theorem?

A line parallel to one side of a triangle divides the other two sides into proportional segments.

What is the converse?

If a line divides two sides of a triangle proportionally, it is parallel to the third side.

How do you solve for a missing segment?

Set the two side ratios equal in a proportion and cross-multiply.

How does the midsegment relate to this theorem?

The midsegment is the special case where the parallel line passes through the midpoints, splitting both sides equally.

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Triangle Proportionality theorem

The Triangle Proportionality Theorem

📖 Theory

How to use the theorem

Common pitfalls

Frequently asked questions

Theory