When can I use AD/DB = AE/EC and when do I need AD/AB = AE/AC?

Both are true when DE BC; they are algebraically equivalent. Use the part-to-part form when the problem gives you the two pieces of a side, and the part-to-whole form when it gives you a piece and the entire side. The only rule is consistency: both fractions in your equation must describe the same kind of relationship.

Why can't I write AD/DB = DE/BC?

Because DE and BC are corresponding sides of the similar triangles ADE and ABC, and the similarity ratio uses whole sides, not pieces. The correct statement is DE/BC = AD/AB = AE/AC. Substituting DB for AB shrinks the denominator and gives a wrong answer for DE.

Does the angle-bisector theorem create similar triangles?

No. Bisecting an angle produces two triangles that share a side and have one pair of congruent angles, but that is not enough for AA, SSS, or SAS similarity. The only guaranteed relationship is BD/DC = AB/AC, matching each piece of the cut side with the adjacent side of the triangle.

How do I know whether lines in a diagram are actually parallel?

You may only assume parallelism if the diagram shows matching arrowheads, the problem statement says so, or you prove it — for example, by showing the two ratios are equal and citing the converse of the triangle proportionality theorem. Appearance is never a justification, and many practice problems are drawn to look parallel when they are not.

GEOM-6.4

Side-Splitter & Proportionality Theorems

Master the triangle proportionality (side-splitter) theorem, its converse, parallel lines on transversals, and the angle-bisector theorem to find missing lengths.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on Side-Splitter & Proportionality Theorems, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

Start Side-Splitter & Proportionality Theorems free

What this lesson covers

You already know that a line parallel to one side of a triangle creates a smaller similar triangle. This lesson turns that fact into a fast tool: the side-splitter theorem, which says the parallel line chops the other two sides into proportional pieces. From there you get three closely related results — the converse (which lets you prove lines are parallel), a version for three parallel lines cut by any two transversals, and the angle-bisector theorem, which splits the opposite side in the ratio of the two adjacent sides.

The hard part is almost never the algebra. It is setting up the correct proportion: deciding whether a length in your diagram is a piece of a side or the whole side. Get that right and every problem in this section becomes a one-step cross-multiplication.

The Side-Splitter (Triangle Proportionality) Theorem

If a line is parallel to one side of a triangle and intersects the other two sides, it divides those two sides proportionally.

In

\triangle ABC

, if

D

is on

\overline{AB}

E

is on

\overline{AC}

, and

\overline{DE} \parallel \overline{BC}

, then

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

Why it works:

\overline{DE} \parallel \overline{BC}

makes

\angle ADE \cong \angle ABC

and

\angle AED \cong \angle ACB

(corresponding angles), so

\triangle ADE \sim \triangle ABC

by AA. Similar triangles give

\frac{AD}{AB} = \frac{AE}{AC}

, and a little algebra converts that whole-to-whole statement into the part-to-part statement above.

That gives you two legitimate but different proportions, and mixing them is the number one source of wrong answers.

Proportion	What the numbers must be
$\frac{AD}{DB} = \frac{AE}{EC}$	piece to piece (top piece over bottom piece)
$\frac{AD}{AB} = \frac{AE}{AC}$	piece to whole side
$\frac{DE}{BC} = \frac{AD}{AB}$	only from similar triangles, never part-to-part

Notice the third row: the segment

\overline{DE}

itself is not part of the part-to-part relationship. If a problem asks for

DE

, you must use the similar-triangle ratio with whole sides, not

\frac{AD}{DB}

. Students who write

\frac{AD}{DB} = \frac{DE}{BC}

get an answer that looks reasonable and is wrong.

A quick habit that prevents this: label each length in the diagram as "upper piece," "lower piece," or "whole," then build the proportion so both sides of the equation describe the same relationship.

The Converse: Proving Two Lines Are Parallel

The side-splitter theorem runs backward. If a line intersects two sides of a triangle and divides them proportionally, then that line is parallel to the third side.

So in

\triangle ABC

with

D

\overline{AB}

and

E

\overline{AC}

, if

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

, you may conclude

\overline{DE} \parallel \overline{BC}

.

This is the tool for justification questions: "Is

\overline{DE}

parallel to

\overline{BC}

? Explain." The procedure is short. Compute both ratios as decimals or reduced fractions, compare them, and state a conclusion with a reason.

Suppose

AD = 9

DB = 6

AE = 12

EC = 8

. Then

\frac{9}{6} = 1.5

and

\frac{12}{8} = 1.5

. The ratios are equal, so by the converse of the triangle proportionality theorem,

\overline{DE} \parallel \overline{BC}

.

Now change

EC

9

. Then

\frac{12}{9} \approx 1.33 \neq 1.5

, so the segments are not parallel. Saying "they look parallel in the picture" is not a justification — diagrams in this unit are frequently drawn to mislead exactly this instinct.

Two cautions. First, both ratios must be written in the same order. Comparing

\frac{AD}{DB}

\frac{EC}{AE}

is comparing a number to its reciprocal, and you will reject a genuinely parallel pair. Second, if a problem gives you a whole side, subtract to get the missing piece before comparing: if

AB = 15

and

AD = 9

, then

DB = 6

, not

15

.

Write your conclusion as a full sentence naming the theorem. That is what a proof-style response needs.

Three Parallel Lines and Two Transversals

The side-splitter idea extends beyond triangles. If three or more parallel lines are cut by two transversals, they divide the transversals proportionally.

Picture parallel lines

\ell

m

, and

n

. A first transversal meets them at

A

B

C

; a second meets them at

X

Y

Z

. Then

\frac{AB}{BC} = \frac{XY}{YZ},

and equivalently

\frac{AB}{XY} = \frac{BC}{YZ}

.

The transversals do not have to intersect anywhere in the picture, and they do not have to be the same length. Only the ratios of corresponding pieces match. This shows up in real settings all the time: parallel street grids cut by two diagonal avenues, evenly spaced fence rails crossed by two slanted braces, or notebook lines used to divide a segment into equal parts.

A special case worth memorizing: if the parallel lines cut congruent segments on one transversal, they cut congruent segments on every transversal. That is the principle behind dividing a segment into

n

equal parts with a straightedge and compass.

Setup errors here are almost always about matching the pieces. The piece between the top two parallel lines on transversal one must be paired with the piece between the same two parallel lines on transversal two. Label the parallel lines top, middle, bottom and tag each segment with which pair of lines bounds it before writing the proportion.

If a problem gives you a total length along one transversal, split it using a variable: if the whole is

20

and the ratio is

2:3

, write the pieces as

2x

and

3x

, solve

5x = 20

, so

x = 4

and the pieces are

8

and

12

The Triangle Angle-Bisector Theorem

This theorem looks like the side-splitter but comes from a different configuration, and confusing the two is a frequent mistake.

If a ray bisects an angle of a triangle, it divides the opposite side into two segments proportional to the other two sides. In

\triangle ABC

, if

\overline{AD}

bisects

\angle BAC

with

D

\overline{BC}

, then

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

Read that carefully: the two pieces of the cut side are compared to the two uncut sides, and each piece is matched with the side it touches at the shared vertex.

BD

sits next to

B

, so it pairs with

AB

;

DC

sits next to

C

, so it pairs with

AC

Situation	Given	Proportion
Line parallel to a side	$\overline{DE} \parallel \overline{BC}$	$\frac{AD}{DB} = \frac{AE}{EC}$
Angle bisector from a vertex	$\overline{AD}$ bisects $\angle A$	$\frac{BD}{DC} = \frac{AB}{AC}$

The angle bisector does not create similar triangles, so do not try to use

\frac{AB}{AC} = \frac{AD}{\text{something}}

. The only guaranteed relationship is the one above.

A typical problem gives

AB = 10

AC = 15

, and

BC = 20

, and asks for

BD

and

DC

. Since the two pieces sum to

20

, let

BD = x

and

DC = 20 - x

. Then

\frac{x}{20-x} = \frac{10}{15} = \frac{2}{3}

, so

3x = 40 - 2x

, giving

x = 8

. Thus

BD = 8

and

DC = 12

. Sanity check: the longer piece,

12

, is adjacent to the longer side,

15

. That check catches a flipped proportion instantly.

Choosing the Right Theorem and Avoiding Setup Errors

Every problem in this lesson is solved by finding one correct proportion. Use the diagram to decide which one.

Ask first: is there a parallel mark? If a segment inside a triangle is parallel to a side, use side-splitter. If three or more parallel lines are crossed by two lines, use the transversal version. If there is an angle bisector (tick marks on two angles at one vertex), use the angle-bisector theorem. If none of these appear but two triangles are marked similar, fall back on corresponding sides from the similarity work earlier in this unit.

The recurring errors are predictable. Students substitute a whole side where a piece belongs, because the diagram labels

AB = 21

while the proportion needs

DB

. Always subtract first. Students also cross-multiply correctly but solve carelessly when the variable appears on both sides, as it does whenever a total length is split into

x

and

\text{total} - x

; expand fully before collecting terms.

A third error is assuming parallelism from appearance. Unless a diagram has arrowheads, a stated parallel condition, or proven equal ratios, you cannot use side-splitter at all.

Finally, check your answer for reasonableness. In side-splitter problems, the larger piece on one side corresponds to the larger piece on the other. In angle-bisector problems, the longer piece of the cut side touches the longer of the two remaining sides. If your solution violates either pattern, you flipped a ratio somewhere — go back to the setup rather than the arithmetic.

Key terms

Side-Splitter (Triangle Proportionality) Theorem.: If a line parallel to one side of a triangle intersects the other two sides, it divides them proportionally: $\frac{AD}{DB} = \frac{AE}{EC}$ .
Converse of the Triangle Proportionality Theorem.: If a line divides two sides of a triangle proportionally, then it is parallel to the third side. Used to justify that two segments are parallel.
Transversal.: A line that intersects two or more other lines at distinct points; in this lesson, the lines cutting across a set of parallel lines.
Three Parallel Lines Theorem.: If three or more parallel lines are cut by two transversals, they divide the transversals proportionally: $\frac{AB}{BC} = \frac{XY}{YZ}$ .
Triangle Angle-Bisector Theorem.: An angle bisector of a triangle divides the opposite side into segments proportional to the two adjacent sides: $\frac{BD}{DC} = \frac{AB}{AC}$ .
Proportion.: An equation stating two ratios are equal, such as $\frac{a}{b} = \frac{c}{d}$ ; solved by cross-multiplying to get $ad = bc$ .
Corresponding parts.: Segments or angles that occupy matching positions in two figures or in a proportional setup; proportions are only valid when parts correspond in the same order.

Worked example

\triangle ABC

, point

D

lies on

\overline{AB}

and point

E

lies on

\overline{AC}

, with

\overline{DE} \parallel \overline{BC}

. You are given

AD = 8

DB = 6

, and

AE = 12

. (a) Find

EC

. (b) Find

AC

. (c) If instead

AE = 12

and

EC = 10

, is

\overline{DE}

still parallel to

\overline{BC}

Part (a). Because

\overline{DE} \parallel \overline{BC}

, the side-splitter theorem applies, so the pieces of

\overline{AB}

are proportional to the pieces of

\overline{AC}

\frac{AD}{DB} = \frac{AE}{EC} \quad \Rightarrow \quad \frac{8}{6} = \frac{12}{EC}.

Cross-multiply:

8 \cdot EC = 6 \cdot 12 = 72

, so

EC = 9

.

Check the pattern:

\frac{8}{6} = \frac{4}{3}

and

\frac{12}{9} = \frac{4}{3}

. The ratios match, so the answer is consistent.

Part (b).

AC

is the whole side, so add the pieces:

AC = AE + EC = 12 + 9 = 21

. A common error here is to report

9

AC

;

9

is only the lower piece. As a second check,

\frac{AD}{AB} = \frac{8}{14} = \frac{4}{7}

and

\frac{AE}{AC} = \frac{12}{21} = \frac{4}{7}

, confirming the piece-to-whole version of the theorem holds.

Part (c). Now test the converse. Compute both part-to-part ratios in the same order:

\frac{AD}{DB} = \frac{8}{6} \approx 1.33, \qquad \frac{AE}{EC} = \frac{12}{10} = 1.2.

Since

1.33 \neq 1.2

, the sides are not divided proportionally, so by the converse of the triangle proportionality theorem

\overline{DE}

is not parallel to

\overline{BC}

— even if the drawing makes it look that way.

Practice questions

Three parallel lines are cut by two transversals. On the first transversal the parallel lines cut off segments of length

4

and

10

, in that order from top to bottom. On the second transversal, the top segment has length

6

and the bottom segment has length

x

. What is

x

$x = 2.4$
$x = 12$
$x = 15$
$x = 16$

Answer: $x = 15$

Three parallel lines cut two transversals proportionally, so

\frac{4}{10} = \frac{6}{x}

. Cross-multiplying gives

4x = 60

, so

x = 15

. Check the reasonableness: on the first transversal the bottom piece is

2.5

times the top piece, and

6 \times 2.5 = 15

, which matches. The value

12

comes from mistakenly adding the difference (

10 - 4 = 6

, then

6 + 6

), and

2.4

comes from flipping the proportion to

\frac{4}{10} = \frac{x}{6}

— a reminder to keep top pieces paired with top pieces on both sides of the equation.

\triangle PQR

, ray

\overline{PS}

bisects

\angle QPR

, with

S

\overline{QR}

. Given

PQ = 9

PR = 12

, and

QR = 14

, find

QS

and

SR

. Show your setup and explain how you know your answer is reasonable.

Answer: $QS = 6$ and $SR = 8$

By the triangle angle-bisector theorem,

\frac{QS}{SR} = \frac{PQ}{PR} = \frac{9}{12} = \frac{3}{4}

. Because

S

lies on

\overline{QR}

, the two pieces must total

14

, so let

QS = x

and

SR = 14 - x

. Then

\frac{x}{14-x} = \frac{3}{4}

, giving

4x = 3(14 - x) = 42 - 3x

, so

7x = 42

and

x = 6

. Therefore

QS = 6

and

SR = 14 - 6 = 8

. The answer is reasonable because the longer piece,

8

, is adjacent to the longer of the two other sides,

PR = 12

, and because

\frac{6}{8} = \frac{3}{4}

matches the required ratio. A frequent slip is pairing

QS

with

PR

instead of

PQ

, which yields

8

and

6

reversed.

\triangle ABC

D

is on

\overline{AB}

with

AD = 10

and

AB = 25

, and

E

is on

\overline{AC}

with

AE = 14

and

EC = 21

. Is

\overline{DE} \parallel \overline{BC}

? Justify your answer.

Answer: Yes, $\overline{DE} \parallel \overline{BC}$ by the converse of the triangle proportionality theorem.

First convert the whole side into a piece:

DB = AB - AD = 25 - 10 = 15

. Now compare the part-to-part ratios in the same order:

\frac{AD}{DB} = \frac{10}{15} = \frac{2}{3}

and

\frac{AE}{EC} = \frac{14}{21} = \frac{2}{3}

. The two ratios are equal, so the line divides sides

\overline{AB}

and

\overline{AC}

proportionally, and by the converse of the triangle proportionality theorem

\overline{DE} \parallel \overline{BC}

. The step students most often skip is the subtraction; comparing

\frac{10}{25}

\frac{14}{21}

mixes a piece-to-whole ratio with a piece-to-piece ratio and would wrongly suggest the segments are not parallel.

FAQ

When can I use $\frac{AD}{DB} = \frac{AE}{EC}$ and when do I need $\frac{AD}{AB} = \frac{AE}{AC}$ ?: Both are true when $\overline{DE} \parallel \overline{BC}$ ; they are algebraically equivalent. Use the part-to-part form when the problem gives you the two pieces of a side, and the part-to-whole form when it gives you a piece and the entire side. The only rule is consistency: both fractions in your equation must describe the same kind of relationship.
Why can't I write $\frac{AD}{DB} = \frac{DE}{BC}$ ?: Because $\overline{DE}$ and $\overline{BC}$ are corresponding sides of the similar triangles $\triangle ADE$ and $\triangle ABC$ , and the similarity ratio uses whole sides, not pieces. The correct statement is $\frac{DE}{BC} = \frac{AD}{AB} = \frac{AE}{AC}$ . Substituting $DB$ for $AB$ shrinks the denominator and gives a wrong answer for $DE$ .
Does the angle-bisector theorem create similar triangles?: No. Bisecting an angle produces two triangles that share a side and have one pair of congruent angles, but that is not enough for AA, SSS, or SAS similarity. The only guaranteed relationship is $\frac{BD}{DC} = \frac{AB}{AC}$ , matching each piece of the cut side with the adjacent side of the triangle.
How do I know whether lines in a diagram are actually parallel?: You may only assume parallelism if the diagram shows matching arrowheads, the problem statement says so, or you prove it — for example, by showing the two ratios are equal and citing the converse of the triangle proportionality theorem. Appearance is never a justification, and many practice problems are drawn to look parallel when they are not.

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