Math Core

When a segment runs parallel to one side of a triangle, it cuts the other two sides into pieces that are in the same ratio. This fact, the Triangle Proportionality Theorem (sometimes called the Side-Splitter Theorem), lets you work directly with the pieces of the sides without setting up two full triangles every time.

The theorem

You already know that if $\overline{DE} \parallel \overline{BC}$ , then $\triangle ADE \sim \triangle ABC$ by AA. That similarity compares the small triangle with the whole triangle. The Triangle Proportionality Theorem goes one step further and compares the two pieces of each side.

Segment DE is parallel to side BC.

Triangle Proportionality Theorem

If a line parallel to one side of a triangle intersects the other two sides, then it divides those two sides proportionally. In the figure, if $\overline{DE} \parallel \overline{BC}$ , then

$\frac{AD}{DB} = \frac{AE}{EC}.$

Why it's true

Here is a proof. It uses the similar triangles and then a little algebra with fractions.

Given: $\overline{DE} \parallel \overline{BC}$ , with $D$ on $\overline{AB}$ and $E$ on $\overline{AC}$ . Prove: $\dfrac{AD}{DB} = \dfrac{AE}{EC}$ .

#	statement	reason
1	$\overline{DE} \parallel \overline{BC}$	Given
2	$\triangle ADE \sim \triangle ABC$	AA Similarity (corresponding angles and the shared $\angle A$ )
3	$\dfrac{AB}{AD} = \dfrac{AC}{AE}$	Corresponding sides of similar triangles are proportional
4	$AB = AD + DB$ , $AC = AE + EC$	Segment Addition Postulate
5	$\dfrac{AD + DB}{AD} = \dfrac{AE + EC}{AE}$	Substitution Property of Equality
6	$1 + \dfrac{DB}{AD} = 1 + \dfrac{EC}{AE}$	Split each fraction
7	$\dfrac{DB}{AD} = \dfrac{EC}{AE}$	Subtraction Property of Equality
8	$\dfrac{AD}{DB} = \dfrac{AE}{EC}$	Take reciprocals of both sides

Because of step 8, you can write the proportion in any consistent way: $\dfrac{AD}{DB} = \dfrac{AE}{EC}$ , or $\dfrac{DB}{AD} = \dfrac{EC}{AE}$ , or $\dfrac{AD}{AB} = \dfrac{AE}{AC}$ (piece over whole).

Worked example: Finding a piece of a side

In $\triangle ABC$ , $\overline{DE} \parallel \overline{BC}$ . If $AD = 4$ , $DB = 6$ and $AE = 5$ , find $EC$ .

$\frac{AD}{DB} = \frac{AE}{EC} \implies \frac{4}{6} = \frac{5}{EC} \implies 4 \cdot EC = 30 \implies EC = 7.5.$

Common mistake

The segment $DE$ is not part of the side-splitter proportion. $\dfrac{AD}{DB} = \dfrac{DE}{BC}$ is wrong. To find $DE$ , use the similar triangles, which compare with the whole side: $\dfrac{DE}{BC} = \dfrac{AD}{AB}$ .

Worked example: Solving for a variable

In $\triangle ABC$ , $\overline{DE} \parallel \overline{BC}$ , $AD = 8$ , $DB = x$ , $AE = 10$ and $EC = x + 2$ . Find $x$ .

$\frac{8}{x} = \frac{10}{x + 2} \implies 8(x + 2) = 10x \implies 8x + 16 = 10x \implies x = 8.$

Check: $DB = 8$ and $EC = 10$ , and $\dfrac{8}{8} = \dfrac{10}{10} = 1$ . Here $D$ and $E$ are midpoints.

That last example connects to the Triangle Midsegment Theorem you learned earlier: a segment joining the midpoints of two sides is parallel to the third side and half as long. The midsegment is just the special case where the ratio is $1 : 1$ .

The converse

The theorem also works in reverse, which gives you a way to prove that segments are parallel using only lengths.

Converse of the Triangle Proportionality Theorem

If a line divides two sides of a triangle proportionally, then it is parallel to the third side. If $\dfrac{AD}{DB} = \dfrac{AE}{EC}$ , then $\overline{DE} \parallel \overline{BC}$ .

Worked example: Is it parallel?

In $\triangle ABC$ , $D$ is on $\overline{AB}$ and $E$ is on $\overline{AC}$ .

a. $AD = 6$ , $DB = 4$ , $AE = 9$ , $EC = 6$ . Compare: $\dfrac{6}{4} = 1.5$ and $\dfrac{9}{6} = 1.5$ . The sides are divided proportionally, so $\overline{DE} \parallel \overline{BC}$ .

b. $AD = 6$ , $DB = 4$ , $AE = 8$ , $EC = 5$ . Compare: $\dfrac{6}{4} = 1.5$ but $\dfrac{8}{5} = 1.6$ . Not proportional, so $\overline{DE}$ is not parallel to $\overline{BC}$ .

Three or more parallel lines

The same idea extends beyond triangles. When parallel lines cross two transversals, they cut both transversals in the same ratio.

Parallel lines and transversals

If three or more parallel lines intersect two transversals, then they divide the transversals proportionally.

Three parallel lines cut two transversals.

In the figure, the parallel lines cut the left transversal into pieces $6$ and $9$ , and the right one into $8$ and $x$ . So

$\frac{6}{9} = \frac{8}{x} \implies 6x = 72 \implies x = 12.$

A special case: if parallel lines cut congruent pieces on one transversal, they cut congruent pieces on every transversal.

The Triangle Angle Bisector Theorem

One more proportion shows up in many problems. It isn't about parallel lines, but it splits a side proportionally in a similar way.

Triangle Angle Bisector Theorem

An angle bisector of a triangle divides the opposite side into two segments proportional to the other two sides. If $\overrightarrow{AD}$ bisects $\angle A$ in $\triangle ABC$ , with $D$ on $\overline{BC}$ , then

$\frac{BD}{DC} = \frac{AB}{AC}.$

Worked example: Splitting a side with a bisector

In $\triangle ABC$ , $AB = 6$ , $AC = 9$ and $BC = 10$ . The bisector of $\angle A$ meets $\overline{BC}$ at $D$ . Find $BD$ and $DC$ .

$\dfrac{BD}{DC} = \dfrac{6}{9} = \dfrac23$ , so $BC$ is split into $2 + 3 = 5$ equal parts of length $\dfrac{10}{5} = 2$ . Then $BD = 2 \cdot 2 = 4$ and $DC = 3 \cdot 2 = 6$ . Check: $4 + 6 = 10$ and $\dfrac46 = \dfrac69$ .

Tip

The shorter piece is always next to the shorter side. That's a quick check for the angle bisector theorem.

Practice

Practice 1

In $\triangle ABC$ , $\overline{DE} \parallel \overline{BC}$ with $D$ on $\overline{AB}$ and $E$ on $\overline{AC}$ . If $AD = 3$ , $DB = 9$ and $AE = 4$ , find $EC$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 2

In $\triangle ABC$ , $\overline{DE} \parallel \overline{BC}$ . If $AD = 5$ , $DB = x$ , $AE = 7.5$ and $EC = 9$ , find $x$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 3

In $\triangle ABC$ , $D$ is on $\overline{AB}$ and $E$ is on $\overline{AC}$ with $AD = 4$ , $DB = 10$ , $AE = 6$ and $EC = 15$ . Is $\overline{DE} \parallel \overline{BC}$ ?

Practice 4

In $\triangle ABC$ , $D$ is on $\overline{AB}$ and $E$ is on $\overline{AC}$ . For which lengths is $\overline{DE}$ not parallel to $\overline{BC}$ ?

Practice 5

In $\triangle ABC$ , $\overline{DE} \parallel \overline{BC}$ with $D$ on $\overline{AB}$ and $E$ on $\overline{AC}$ . If $AD = 6$ , $DB = 3$ and $BC = 12$ , find $DE$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 6

In $\triangle ABC$ , $\overline{DE} \parallel \overline{BC}$ , $AD = x$ , $DB = 6$ , $AE = x + 2$ and $EC = 9$ . Find $x$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 7

Three parallel lines cross two transversals. On the first transversal they cut pieces of length $4$ and $10$ . On the second, the piece matching the $4$ has length $6$ . How long is the piece matching the $10$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 8

In $\triangle ABC$ , $AB = 8$ , $AC = 12$ and $BC = 15$ . The bisector of $\angle A$ meets $\overline{BC}$ at $D$ . Find $BD$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.