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Triangle Proportionality theorem

20 practice questions 2 video lessons Theory + worked examples

The Triangle Proportionality Theorem

Texas Geometry (TEKS) • Standard G.8(A) • Similarity

The Triangle Proportionality Theorem is a topic in Similarity in the Texas Essential Knowledge and Skills (Geometry, §111.41). It is aligned to Standard G.8(A), which requires students to prove and apply the proportionality of segments cut by a line parallel to a side of a triangle.

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

Texas Geometry (TEKS) › Similarity › The Triangle Proportionality Theorem  —  Standard G.8(A)

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Practice questions

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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.
Triangle Proportionality Theorem A line parallel to one side of a triangle divides the other two sides proportionally. D E A B C DE ∥ AB divides the sides proportionally
\(DE\parallel AB\) splits the sides proportionally.
Proportionality Proportionality Proportionality DE ∥ AB ⇒ CD / DA = CE / EB
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

  1. Confirm the line is parallel to a side.
  2. Match the two pieces of each split side.
  3. Write the proportion \(\dfrac{CD}{DA}=\dfrac{CE}{EB}\).
  4. 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.