Overview

The Basic Proportionality Theorem (BPT), also called the Side Splitter Theorem or Thales' Theorem, describes what happens when a line parallel to one side of a triangle cuts through the other two sides. It produces proportional segments, and recognising this setup is one of the most common moves in competition geometry.

Theorem Statement

In triangle $ABC$, suppose a line parallel to $BC$ intersects $AB$ at $D$ and $AC$ at $E$. Then

In words, a line parallel to one side of a triangle divides the other two sides in the same ratio.

Equivalent Forms

Starting from $\frac{AD}{DB} = \frac{AE}{EC}$, simple algebra gives several useful restatements. If we set the common ratio equal to $k$, so that $AD = k(DB)$ and $AE = k(EC)$, then:

The last equality ($DE/BC$) follows because triangle $ADE$ is similar to triangle $ABC$ with ratio $k/(k+1)$.

A particularly clean version, if $DE \parallel BC$, then

This form is often the fastest to apply in contest problems.

Why It Works

The standard proof uses areas.

Given $DE \parallel BC$ in triangle $ABC$, with $D$ on $AB$ and $E$ on $AC$.

Construct $DM \perp AC$ and $EN \perp AB$.

Consider triangles $ADE$ and $DBE$, which share the same height $EN$ from $E$ to line $AB$

Similarly, triangles $ADE$ and $DEC$ share the same height $DM$ from $D$ to line $AC$

Now, since $DE \parallel BC$, triangles $DBE$ and $DEC$ sit on the same base $DE$ and between the same pair of parallel lines $DE$ and $BC$. Therefore they have equal area

Combining everything:

The Converse

The converse of BPT is just as useful

If $D$ lies on $AB$ and $E$ lies on $AC$ such that

then $DE \parallel BC$.

In contest problems, this is often how you prove two lines are parallel, show that a transversal cuts two sides of a triangle proportionally.

Connection to Similar Triangles

BPT and triangle similarity are quite linked

If $DE \parallel BC$ in triangle $ABC$, then triangle $ADE$ is similar to triangle $ABC$ (by AA similarity, since $\angle ADE = \angle ABC$ and $\angle AED = \angle ACB$ from the parallel lines).

The ratio of similarity is $AD/AB = AE/AC = DE/BC$.

This means

• All corresponding lengths scale by the same factor.
• Areas scale by the square of that factor.

Special Case: The Midpoint Theorem

When the parallel line passes through the midpoints of two sides, BPT gives a ratio of $1{:}1$, and we get the Midpoint Theorem

The segment joining the midpoints of two sides of a triangle is parallel to the third side and equal to half its length.

If $D$ and $E$ are midpoints of $AB$ and $AC$ respectively, then $DE \parallel BC$ and $DE = \frac{1}{2} BC$.

Trapezoid Diagonals

BPT extends naturally to trapezoids. In trapezoid $ABCD$ with $AB \parallel CD$, the diagonals $AC$ and $BD$ intersect at a point $P$ such that

This follows by applying BPT (or similar triangles) to the triangles formed by the diagonals.

Parallel Lines Cut by Transversals

A useful generalisation, if three or more parallel lines are cut by two transversals, the ratios of corresponding segments on the two transversals are equal.

Given parallel lines $\ell_1, \ell_2, \ell_3$ cutting transversals $t$ and $s$ at points $A, B, C$ and $D, E, F$ respectively

This reduces to BPT by connecting two points to form a triangle and applying the theorem inside it.

Worked Examples

Example 1

In triangle $ABC$, point $D$ lies on $AB$ and point $E$ lies on $AC$ such that $DE \parallel BC$. If $AD = 3$, $DB = 5$, and $AE = 4.5$, find $EC$.

By BPT,

Substituting:

Cross multiplying: $3(EC) = 5(4.5) = 22.5$, so $EC = 7.5$.

Example 2

In triangle $PQR$, points $S$ and $T$ lie on $PQ$ and $PR$ such that $ST \parallel QR$. If $PS = 6$, $PQ = 10$, and $QR = 15$, find $ST$.

Since $ST \parallel QR$, triangle $PST$ is similar to triangle $PQR$ with ratio

Therefore

Example 3 (Area Ratios)

In triangle $ABC$ with area $40$, a line parallel to $BC$ meets $AB$ at $D$ and $AC$ at $E$ such that $AD/AB = 2/5$. Find the area of trapezoid $BCED$.

Since $DE \parallel BC$ and the ratio of similarity is $2/5$, the area of triangle $ADE$ is

The area of trapezoid $BCED$ is $40 - 6.4 = 33.6$.

Example 4 (2018 AMC 10A Problem 9)

All of the triangles in the diagram are similar to isosceles triangle $ABC$ with $AB = AC$. Each of the $7$ smallest triangles has area $1$, and $\triangle ABC$ has area $40$. What is the area of trapezoid $DBCE$?

The $7$ small triangles along the base of the upper region have total area $7$, but the triangle $ADE$ sitting above the trapezoid also includes a larger similar triangle above the row of $7$. By looking at the structure, the base of $ADE$ is $4$ times the base of each small triangle, so

Since each small triangle has area $1$, the area of $\triangle ADE$ is $16$.

Therefore the area of trapezoid $DBCE = 40 - 16 = 24$.

Key Contest Insight

The most common contest pattern is:

1. Spot a line parallel to one side of a triangle (or construct one).
2. Apply BPT to get proportional segments.
3. Use the resulting similar triangles for lengths, areas, or angle chasing.

Area ratios are very useful, if two similar figures have side ratio $k$, their area ratio is $k^2$. Many problems that seem to require coordinates or trigonometry become short once you identify the parallel line and apply BPT.

Strategy Tips

• Whenever you see a trapezoid in a contest problem, look for similar triangles formed by extending the non parallel sides until they meet. BPT often applies in the resulting triangle.
• If a problem gives you parallel lines and asks for a ratio, BPT is almost certainly the intended tool.
• When multiple parallel lines appear, apply BPT repeatedly or use the transversal generalisation.
• The converse is useful for proving parallelism, if you can show two segments divide two sides of a triangle proportionally, the segments are parallel to the third side.

Common Pitfalls

• Confusing which segments are in the ratio. The theorem says $AD/DB = AE/EC$, not $AD/AB = AE/EC$ (though the latter is also true, it is a different ratio).
• Forgetting that BPT requires the line to be parallel to a side. Without parallelism, the segments are not proportional.
• Applying the area squared rule ($k^2$) to lengths instead of areas, or vice versa.
• Missing that the converse can be used to establish parallelism when the problem does not state it directly.

Practice Problems