What does the intercept theorem say?

If a line parallel to one side of a triangle meets the other two sides at D and E, it cuts them in the same ratio: AD/DB = AE/EC, or equivalently AD/AB = AE/AC. It is also called the basic proportionality theorem or the side-splitter theorem.

How is the intercept theorem proved with areas?

Triangles BDE and CDE have the same base DE and equal heights (B and C lie on a parallel to DE), so their areas are equal. Then AD/DB = [ADE]/[BDE] and AE/EC = [ADE]/[CDE] by equal heights; same numerator, equal denominators, so the ratios are equal.

Is the converse of the intercept theorem true?

Yes. If D on AB and E on AC satisfy AD/DB = AE/EC, then DE is parallel to BC. The same areas prove it: the equal ratios force [BDE] = [CDE], so B and C are at the same distance from the line DE.

Intercept Theorem

How to use

▶ Play runs the proof, Next → and ← Back step one scene at a time, Reset restores the opening scene and the original figure. Learn more about the controls
Drag $D$ along $AB$ , or $A$ , $B$ or $C$ ; the readout under the canvas shows $AD/DB$ , $AE/EC$ and $DE/BC$ . Learn more about why it holds
Labels toggles the point names, Measures writes each marked angle's size beside its arc. The Step explanations log keeps every step so far. Learn more about reading a scene

Proof?DE is parallel to BC. Triangles BDE and CDE share the base DE and their apexes B and C lie on a line parallel to it, so they have the same height and the same area. Then AD/DB = [ADE]/[BDE] and AE/EC = [ADE]/[CDE], two fractions with the same numerator and equal denominators. Drag D along AB, or a vertex.Intercept theorem

DisplayLabelsMeasures

Drag D along AB, or A, B or C

AD/DB = 0.82,  AE/EC = 0.82,  DE/BC = 0.45

Step 1 / 5

Step explanations

1A line parallel to one side

Given
In triangle ABC the line DE is parallel to the side BC, with D on AB and E on AC. To prove: AD/DB = AE/EC, the parallel cuts the two sides in the same ratio.

Drag D: the parallel moves, E follows. Learn more about the statement

Getting Started with the Visualizer

The Statement

Equal Areas Between Parallels

Getting Started with the Visualizer

DemoA run through the proof player

Step 0 of 6

Step through the proof, then drag

D

along the side

AB

to move the parallel, or drag a vertex, and watch the two ratios in the readout stay equal. The picture below is the player with its parts numbered; the other tabs list what can be dragged, what each control does and what the step log shows.

Opening scene: the controls

1234567

1Proof — the theorem loaded in the player; the ? gives its idea in one line

2Display — Labels shows the point names, Measures writes each marked angle's size beside its arc

3Draggable points — D slides along AB; A, B and C move freely; the hollow point E follows so that DE stays parallel to BC

4Readout — the line under the figure: AD/DB, AE/EC and DE/BC, live while you drag

5Back, Play, Next, Reset — step through the proof or run it; Reset returns to step 1 and restores the original figure

6Speed and step counter — how fast Play advances, and which of the 5 steps is on screen

7Step explanations — every step reached so far, with its justification tag and a link to the section that explains it; click an entry to jump back to it

Proof complete: step 5 of 5

Proof?Intercept theoremDisplayLabelsMeasuresDrag D along AB, or A, B or C

AD/DB = 0.82,  AE/EC = 0.82,  DE/BC = 0.45

← Back↻ ReplayNext →ResetNormal ▾Step 5 / 5

Step explanations

1A line parallel to one side

Given
In triangle ABC the line DE is parallel to the side BC, with D on AB and E on AC. To prove: AD/DB = AE/EC, the parallel cuts the two sides in the same ratio.

2[BDE] = [CDE]

Parallel lines
Join B to E and C to D. Triangles BDE and CDE have the same base DE, and their apexes B and C lie on the line BC, parallel to DE, so they are at the same distance from it: equal heights on the same base, equal areas.

3AD/DB = [ADE] / [BDE]

Same height
Triangles ADE and BDE have their bases AD and DB on the line AB and the same apex E, so they have the same height, and their areas are in the ratio of their bases.

4AE/EC = [ADE] / [CDE]

Same height
The same on the other side: ADE and CDE have their bases AE and EC on the line AC and the same apex D, so [ADE] / [CDE] = AE/EC.

5AD/DB = AE/EC

Equal denominators
The two fractions have the same numerator [ADE], and their denominators [BDE] and [CDE] are equal. So AD/DB = AE/EC; flipping both sides and adding 1 gives AB/AD = AC/AE as well. ∎

12345

1The parallel — DE, marked with the same chevron as BC

2The shaded triangles — ADE at the top, BDE and CDE below it on the base DE

3The cross lines — BE and CD, the sides of the two equal triangles

4Justification tags — Given, Parallel lines, Same height (twice), Equal denominators

5Replay — at the last step Play becomes Replay and Next is off

The Statement

The intercept theorem, also called the basic proportionality theorem or Thales' theorem in some countries, says: if a line parallel to one side of a triangle meets the other two sides, it cuts them in the same ratio. In triangle

ABC

, with

D

AB

E

AC

and

DE \parallel BC

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

Equivalently

AD/AB = AE/AC

: the small triangle

ADE

has its two sides from

A

in the same ratio to the sides of

ABC

. The theorem is the gateway to similar triangles; the AA similarity page is built on it.

The proof here is by areas, in the same style as Ceva and Menelaus: nothing but the formula half base times height. In the player

D

slides along

AB

and

E

is placed on

AC

so that

DE

stays parallel to

BC

; the readout shows the two ratios and

DE/BC

Opening scene, frozen

DE parallel to BC, with D on AB and E on AC.

Equal Areas Between Parallels

Join

B

E

and

C

D

. The triangles

BDE

and

CDE

stand on the same base

DE

, and their third vertices

B

and

C

lie on the line

BC

, which is parallel to

DE

. Two parallel lines stay the same distance apart, so the heights of the two triangles from

B

and from

C

down to the line

DE

are equal:

[BDE] = [CDE]

This is the one place where the parallel is used. Everything after it is the same-height argument from the angle bisector theorem page.

Equal areas between parallels, frozen

BDE and CDE shaded: same base DE, equal heights, equal areas.

Cutting the Side AB

The triangles

ADE

and

BDE

have their bases

AD

and

DB

on the line

AB

and the same apex

E

, so they have the same height, and their areas are in the ratio of their bases:

\frac{[ADE]}{[BDE]} = \frac{AD}{DB}

Cutting the side AB, frozen

ADE and BDE with the apex E: [ADE]/[BDE] = AD/DB.

Cutting the Side AC

The same on the other side. The triangles

ADE

and

CDE

have their bases

AE

and

EC

on the line

AC

and the same apex

D

\frac{[ADE]}{[CDE]} = \frac{AE}{EC}

Cutting the side AC, frozen

ADE and CDE with the apex D: [ADE]/[CDE] = AE/EC.

Why It Holds

The two fractions have the same numerator

[ADE]

, and their denominators

[BDE]

and

[CDE]

are equal by the parallel. So

\frac{AD}{DB} = \frac{[ADE]}{[BDE]} = \frac{[ADE]}{[CDE]} = \frac{AE}{EC}

Flip both sides and add 1 to get the other common form:

(DB + AD)/AD = (EC + AE)/AE

, that is

AB/AD = AC/AE

. The end-of-proof mark ∎ in the log closes the argument. The tags name the facts used: the given figure, equal areas between parallels, same-height triangles twice, and equal denominators.

The converse is also true: if

AD/DB = AE/EC

then

DE \parallel BC

. It has the same proof read backwards, and the SAS similarity page carries it out.

The conclusion, frozen

Same numerator, equal denominators: AD/DB = AE/EC.

Dragging the Figure

Drag

D

along

AB

: the parallel

DE

moves up and down the triangle,

E

slides along

AC

to keep it parallel, and the two ratios in the readout stay equal while they change together. The third number,

DE/BC

, equals

AD/AB

; that is the next theorem, on the AA similarity page. Drag

A

B

C

to reshape the triangle;

D

keeps its fraction of

AB

.

The player keeps

D

between

A

and

B

. The theorem also holds when the parallel meets the extensions of the sides beyond

A

or beyond

B

and

C

, with the same proof, but the player keeps to the case where the parallel crosses the triangle. A drag that would carry a vertex out of the frame or flatten the triangle is refused.

After dragging D, frozen

D further down AB: both ratios grow together, and DE/BC with them.

Consequences

• Similar triangles. The theorem is the first step of the AA similarity criterion: cut off a copy of the smaller triangle at the vertex

A

, and the intercept theorem gives the proportions.
• The midsegment theorem again. When

D

is the midpoint of

AB

, the equal ratios force

E

to be the midpoint of

AC

; that is the converse direction of the midsegment theorem.
• Dividing a segment. To cut a segment into

n

equal parts with a ruler and compass, draw any ray from one end, mark

n

equal steps on it, join the last mark to the other end and draw parallels through the other marks; the intercept theorem makes the pieces equal.
• The converse. If a line through

D

and

E

cuts

AB

and

AC

in the same ratio, it is parallel to

BC

; the SAS similarity page proves this by the same areas.
• Ceva and Menelaus. Ceva and Menelaus use the same same-height argument on the three sides of a triangle at once.