What does the midsegment theorem say?

The segment joining the midpoints of two sides of a triangle is parallel to the third side and half as long.

Can the midsegment theorem be proved without similar triangles?

Yes. Extend the midsegment to double its length and join the end to the third vertex; two side-angle-side congruences, with vertical angles and alternate angles, show that the doubled segment and the third side are the opposite sides of a parallelogram.

What is the medial triangle?

The triangle formed by the three midsegments. Its sides are parallel to the sides of the original triangle and half as long, and it cuts the original into four congruent triangles.

The Midsegment 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 $A$ , $B$ or $C$ ; the midpoints and the construction point $F$ follow, and the readout under the canvas keeps $DE$ at half of $BC$ . 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?Extend DE past E to F with EF = DE and join CF. Triangles ADE and CFE are congruent (SAS, with the vertical angles at E), so CF = AD = DB and the alternate angles at A and C make CF ∥ AB. Then DBCF has one pair of opposite sides equal and parallel; the diagonal DC and a second SAS show it is a parallelogram, so DF = BC and DF ∥ BC: DE is half of BC and parallel to it. Drag any vertex.Midsegment theorem

DisplayLabelsMeasures

Drag A, B or C; the midpoints and F follow

DE = 2.50, BC = 5.00  (BC / 2 = 2.50)

Step 1 / 8

Step explanations

1The midpoints of two sides

Given
In triangle ABC, D is the midpoint of AB (one tick each) and E the midpoint of AC (two ticks each). To prove: DE ∥ BC and DE = ½ BC.

D, E are computed midpoints; drag a vertex and they follow. Learn more about the statement

Getting Started with the Visualizer

The Statement

The Extension

Vertical Angles and SAS

A Parallel Appears

Three Equal Segments

The Second Transversal

The Second Congruence

Why It Holds

Dragging the Figure

Consequences

Getting Started with the Visualizer

DemoA run through the proof player

Step 0 of 9

Step through the proof, then drag the vertices and watch the segment between the midpoints stay parallel to the third side and half its length. 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 — A, B and C move freely; the hollow midpoints D, E and the construction point F follow

4Readout — the line under the figure: DE and BC in the frame's units, with half of 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 8 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 8 of 8

Proof?Midsegment theoremDisplayLabelsMeasuresDrag A, B or C; the midpoints and F follow

DE = 2.50, BC = 5.00  (BC / 2 = 2.50)

← Back↻ ReplayNext →ResetNormal ▾Step 8 / 8

Step explanations

1The midpoints of two sides

Given
In triangle ABC, D is the midpoint of AB (one tick each) and E the midpoint of AC (two ticks each). To prove: DE ∥ BC and DE = ½ BC.

2Extend DE to F with EF = DE, and join CF

Construction
Draw DE on past E to the point F with EF = DE (three ticks each), and join C to F. The point F is hollow: it follows D and E.

3Triangles ADE and CFE are congruent

SAS congruence
AE = EC (E is a midpoint), DE = EF (by construction), and the angles at E between those sides are vertical angles, so they are equal. Two sides and the included angle: the triangles are congruent, with A ↔ C, D ↔ F, E ↔ E.

4CF ∥ AB

Alternate angles
Corresponding angles of the congruent triangles are equal: ∠DAE = ∠FCE. These are alternate angles on the transversal AC of the lines AB and CF, and equal alternate angles mean the lines are parallel.

5CF = AD = DB

Corresponding parts
Corresponding sides of the congruent triangles are equal: CF = AD. And AD = DB because D is the midpoint of AB. So CF = DB (one tick each): DBCF has a pair of opposite sides that are equal and parallel.

6The diagonal DC: ∠BDC = ∠FCD

Alternate angles
Join D to C. DC is a transversal of the parallels DB and CF, so the alternate angles ∠BDC and ∠FCD are equal.

7Triangles DBC and CFD are congruent

SAS congruence
DB = CF, DC is a side of both, and the angles between those sides, at D and at C, were just shown equal. So the triangles are congruent, with D ↔ C, B ↔ F, C ↔ D: BC = FD, and ∠BCD = ∠FDC.

8DE ∥ BC and DE = ½ BC

Corresponding parts
∠BCD = ∠FDC are alternate angles on the transversal DC, so DF ∥ BC, and DE lies along DF. And BC = DF = DE + EF = 2 DE, so DE = ½ BC. ∎

12345

1The midsegment DE — parallel to BC (one chevron each) and half as long

2The extension to F — DE doubled; CF = DB and CF ∥ DB make DBCF a parallelogram

3The two congruences — ADE with CFE at the top, DBC with CFD across the diagonal DC

4Justification tags — Given, Construction, SAS congruence, Alternate angles, Corresponding parts, Alternate angles, SAS congruence, Corresponding parts

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

The Statement

Take the midpoints

D

and

E

of two sides

AB

and

AC

of a triangle. The segment

DE

joining them is called a midsegment (or midline) of the triangle, and the theorem says two things about it:

DE \parallel BC, \qquad DE = \tfrac{1}{2} BC

it is parallel to the third side and half as long. The theorem is usually met together with similar triangles, where it is one line: triangle

ADE

is a half-size copy of triangle

ABC

. The proof on this page needs no similarity. It uses only congruent triangles and parallel lines, the tools of the earlier pages, and it is a little longer for that: two congruences instead of one scaling.

In the player

D

E

and the construction point

F

are hollow, computed from the vertices. The readout under the canvas shows

DE

and

BC

in the frame's own units, with half of

BC

beside them, live while you drag.

Opening scene, frozen

Triangle ABC with the midpoints D of AB and E of AC, and the midsegment DE.

The Extension

The construction doubles the midsegment. Draw

DE

on past

E

to the point

F

with

EF = DE

, so that

E

is the midpoint of

DF

as well as of

AC

; and join

C

F

. The player marks

DE

and

EF

with three ticks each.

The point

E

is now the midpoint of two segments at once,

AC

and

DF

, which cross there. That is the whole idea: two segments bisecting each other are the diagonals of a parallelogram, as the parallelogram diagonals page shows in one direction and this proof shows in the other, so

ADCF

is a parallelogram and

CF

is a copy of

AD

, moved across. The next three steps say this with congruent triangles.

The extension, frozen

DE doubled to F (three ticks each), and CF joined.

Vertical Angles and SAS

Look at the two triangles

ADE

and

CFE

that meet at

E

AE = EC

, because

E

is the midpoint of

AC

DE = EF

, by construction. And the angles at

E

, between

EA

and

ED

in one triangle and between

EC

and

EF

in the other, are vertical angles, so they are equal; the player calls both

\varepsilon

.

Two sides and the included angle: the triangles are congruent by side-angle-side, with the vertices matched

A \leftrightarrow C

D \leftrightarrow F

E \leftrightarrow E

. The second triangle is the first one turned through half a turn about

E

The first congruence, frozen

ADE and CFE: AE = EC, DE = EF, vertical angles ε at E.

A Parallel Appears

Corresponding angles of congruent triangles are equal, so

\angle DAE = \angle FCE

; the player marks both

\alpha

. Now look at where these angles sit: one at

A

between

AB

and

AC

, the other at

C

between

CF

and

CA

. They are alternate angles on the transversal

AC

of the two lines

AB

and

CF

. Equal alternate angles mean parallel lines, by the converse of the parallel lines theorem:

CF \parallel AB

The player marks

BA

and

CF

with two chevrons each, pointing the same way.

A parallel appears, frozen

α at A and at C are alternate angles on AC: CF ∥ AB (two chevrons each).

Three Equal Segments

Corresponding sides of the congruent triangles are equal too:

CF = AD

. And

AD = DB

, because

D

is the midpoint of

AB

. So

CF = AD = DB

and the player marks

CF

with one tick, like the two halves of

AB

. Put the last two steps together: the quadrilateral

DBCF

has the pair of opposite sides

DB

and

CF

equal and parallel. That is enough to make it a parallelogram, and the next two steps prove it, in the same way the opposite sides page proved its theorem: with a diagonal and a congruence.

Three equal segments, frozen

CF = AD = DB, one tick each.

The Second Transversal

Join

D

C

. The segment

DC

is a diagonal of

DBCF

, and it is a transversal of the parallel lines

DB

and

CF

. The angle

\angle BDC

D

and the angle

\angle FCD

C

lie on opposite sides of the transversal, between the parallels, so they are alternate angles and are equal; the player marks both

\gamma

.

This is the parallel lines theorem used in its forward direction, from parallel lines to equal angles, after the previous step used it backwards, from equal angles to parallel lines.

The second transversal, frozen

DC joined; γ at D and at C are alternate angles on it.

The Second Congruence

Now compare the triangles

DBC

and

CFD

, on either side of the diagonal.

DB = CF

from Three Equal Segments;

DC

is a side of both; and the angles between those sides,

\gamma

D

and

\gamma

C

, are equal. Side-angle-side again: the triangles are congruent, with

D \leftrightarrow C

B \leftrightarrow F

C \leftrightarrow D

.

Two things follow. The side

BC

corresponds to

FD

, so

BC = DF

. And the angle

\angle BCD

corresponds to

\angle FDC

, so they are equal; the player marks both

\delta

. The argument is exactly the one on the opposite sides page, run with one pair of parallels known and the other to be found.

The second congruence, frozen

DBC and CFD: DB = CF, DC shared, γ between; so BC = FD and the angles δ are equal.

Why It Holds

The two equal angles

\delta

are alternate angles on the transversal

DC

of the lines

DF

and

BC

, so

DF \parallel BC

; and

DE

is part of

DF

, so

DE \parallel BC

. For the length,

BC = DF = DE + EF = 2 \cdot DE

, so

DE = \tfrac{1}{2} BC

. The end-of-proof mark ∎ in the log closes the argument. The tags name the facts used: the given midpoints, a construction, side-angle-side with vertical angles, alternate angles, corresponding parts, alternate angles again, side-angle-side again, and corresponding parts.

Stripped of its two congruences the proof says:

ADCF

is a parallelogram because its diagonals bisect each other, so

CF

equals and is parallel to

AD

, hence to

DB

; then

DBCF

is a parallelogram because one pair of its opposite sides is equal and parallel, so

DF

equals and is parallel to

BC

. The congruences are what those two "because"s cost when written out.

The conclusion, frozen

DF ∥ BC (one chevron each) and BC = 2 DE.

Dragging the Figure

Drag any vertex: the midpoints slide along their sides, the midsegment turns to stay parallel to

BC

, the construction point

F

follows, and the readout keeps

DE

at exactly half of

BC

. Drag

A

sideways and watch the parallelogram

DBCF

shear; drag

B

C

and watch

DE

grow and shrink with the base.

A drag that would flatten the triangle, or carry a vertex or the point

F

out of the frame, is refused. The theorem itself has no such limits; they are the frame's.

After dragging A, frozen

A pulled right: the triangle leans and the parallelogram DBCF shears; DE is still half of BC and parallel to it.

Consequences

• The medial triangle. The three midsegments of a triangle cut it into four congruent triangles, each a half-size copy of the original, so the medial triangle has a quarter of the area and its sides are parallel to the original's.
• Varignon's theorem. Joining the midpoints of the sides of any quadrilateral gives a parallelogram: each pair of opposite sides of the new figure is a midsegment of a triangle on one diagonal, so both are parallel to that diagonal and half its length.
• Trapezoids. The segment joining the midpoints of the legs of a trapezoid is parallel to the bases and equal to half their sum; it comes from this theorem applied along a diagonal.
• The converse. A line through the midpoint of one side of a triangle, parallel to a second side, bisects the third side.
• Angles. Since

DE \parallel BC

, the angles of triangle

ADE

D

and

E

equal the angles of

ABC

B

and

C

(corresponding angles), and its angle sum is the same

180°

; this is the beginning of the similarity of the two triangles.