Triangle Basic Proportionality Theorem

Last updated on 17 November 2025

Table of contents

What does Basic Proportionality theorem state?
Proof of Basic Proportionality Theorem
Converse of Basic Proportionality theorem
FAQs on Triangle Basic Proportionality Theorem

Basic Proportionality Theorem, also known as the Thales' Theorem or Side Splitter Theorem, is a fundamental principle in geometry attributed to the ancient Greek mathematician Thales of Miletus. The theorem deals with the proportionality of line segments in triangles and parallel lines. This theorem forms the basis for various geometric and proportional relationships, playing a key role in understanding parallel lines and their intersections with triangles. Thales' Theorem has applications in diverse fields, including engineering, architecture, and physics, where understanding proportionality is essential for accurate design and analysis.

What does Basic Proportionality theorem state?

This theorem states that if a line is drawn parallel to one side of a triangle and it intersects the other two sides of the triangle, then it divides those two sides proportionally.

Observe △ABC in above figure, Line PQ intersects side AB and side AC at points P and Q respectively and it is parallel to side BC, hence by Basic Proportionality theorem,

\frac{AP}{PB} = \frac{AQ}{QC}

Proof of Basic Proportionality Theorem

Now let's prove the Basic Proportionality Theorem.

Given:

In △ABC in the figure above,

Line PQ intersects side AB and side AC at points P and Q respectively.
Line PQ is parallel to side BC i.e. $PQ ∥ BC$ .

Construction:

Drawn line segment QX perpendicular to side AB.
Drawn line segment PY perpendicular to side AC.
Joined the vertex B of ΔABC to point Q and the vertex C to point P to form the lines BQ and CP.

To Prove: $\frac{AP}{PB} = \frac{AQ}{QC}$

Proof:

As we know that,

Area of triangle = \frac{1}{2} \times base \times height

\begin{aligned} (1) & ∴ & A (△ APQ) & = & \frac{1}{2} \times AP \times QX \\ (2) & ∴ & A (△ PQB) & = & \frac{1}{2} \times PB \times QX \\ (3) & ∴ & A (△ APQ) & = & \frac{1}{2} \times AQ \times PY \\ (4) & ∴ & A (△ QPC) & = & \frac{1}{2} \times QC \times PY \end{aligned}

Divide equation $(1)$ by $(2)$ .

\begin{aligned} \frac{A (△ APQ)}{A (△ PQB)} & = & \frac{\frac{1}{2} \times AP \times QX}{\frac{1}{2} \times PB \times QX} \\ (5) & \frac{A (△ APQ)}{A (△ PQB)} & = & \frac{AP}{PB} \end{aligned}

Now, divide equation $(3)$ by $(4)$ .

\begin{aligned} \frac{A (△ APQ)}{A (△ QPC)} & = & \frac{\frac{1}{2} \times AQ \times PY}{\frac{1}{2} \times QC \times PY} \\ (6) & \frac{A (△ APQ)}{A (△ QPC)} & = & \frac{AQ}{QC} \end{aligned}

According to the property of triangles, the triangles drawn between the same parallel lines and on the same base have equal areas.
Since, ΔPQB and ΔQPC are on the same base PQ, and between the same parallel lines PQ and BC, we can say that ΔPQB and ΔQPC have the same area. $A (△ PQB) = A (△ QPC)$ Therefore, $\begin{aligned} \frac{△ APQ}{△ PQB} & = & \frac{△ APQ}{△ QPC} \\ From (5) and (6) & \frac{AP}{PB} & = & \frac{AQ}{QC} \end{aligned}$

Conclusion: Hence we proved the Basic Proportionality theorem. Therefore, the line PQ drawn parallel to the side BC of ΔABC divides the other two sides AB and AC in equal proportion.

Converse of Basic Proportionality theorem

Converse of the Basic Proportionality theorem states that if a line intersecting two sides of a triangle in two distinct points divides those sides in same ratio, then the line is parallel to the third side of the triangle.

Observe △ABC in the figure above, Line PQ intersects side AB and side AC at points P and Q respectively. If this line divides the sides AB and AC in the same ratio, i.e. if

\frac{AP}{PB} = \frac{AQ}{QC}

, then by Converse of Basic Proportionality theorem, line PQ is parallel to side BC i.e.

PQ ∥ BC

FAQs on Triangle Basic Proportionality Theorem

How can Triangle Basic Proportionality Theorem be expressed mathematically?
If a line segment is drawn parallel to one side of triangle ABC, intersecting the other two sides at points D and E, then:
$\frac{AD}{DB} = \frac{AE}{EC}$
What are the conditions for the Triangle Basic Proportionality Theorem to hold?
The theorem applies in any triangle when a line is drawn parallel to one of its sides, creating two proportional segments on the other two sides.
Is the Triangle Basic Proportionality Theorem applicable in similar triangles?
Yes, the theorem is closely related to the properties of similar triangles, since similar triangles have corresponding sides that are proportional.
How can this theorem be used in real-life applications?
The Triangle Basic Proportionality Theorem is used in various fields such as architecture, engineering, and graphics, where determining proportions is essential.
Can the Triangle Basic Proportionality Theorem be used to find lengths?
Yes! You can set up a proportion based on the segments created by the parallel line to solve for unknown lengths using cross-multiplication.
What if the line is not exactly parallel?
If the line is not parallel to one side of the triangle, the segments formed will not follow the proportional relationship stated in the theorem.
Are there any exceptions to the Triangle Basic Proportionality Theorem?
No exceptions exist as long as the conditions of the theorem (a line parallel to one side of a triangle) are met.

Triangle Basic Proportionality Theorem

What does Basic Proportionality theorem state?

Proof of Basic Proportionality Theorem

Converse of Basic Proportionality theorem

FAQs on Triangle Basic Proportionality Theorem

How can Triangle Basic Proportionality Theorem be expressed mathematically?

What are the conditions for the Triangle Basic Proportionality Theorem to hold?

Is the Triangle Basic Proportionality Theorem applicable in similar triangles?

How can this theorem be used in real-life applications?

Can the Triangle Basic Proportionality Theorem be used to find lengths?

What if the line is not exactly parallel?

Are there any exceptions to the Triangle Basic Proportionality Theorem?

Triangle Angle Sum Theorem

Triangle Exterior Angle Theorem