What does Varignon's theorem say?

The midpoints of the four sides of any quadrilateral are the vertices of a parallelogram, whose sides are parallel to the diagonals of the quadrilateral and half as long.

Why is the midpoint figure a parallelogram?

Each diagonal cuts the quadrilateral into two triangles, and two opposite sides of the midpoint figure are midsegments of those triangles, so both are parallel to that diagonal. With the other diagonal the other pair is parallel too.

Does it work for a non-convex quadrilateral?

Yes. The proof only uses the midsegment theorem inside the four triangles cut off by the diagonals, which exist for any four points, convex or not, even in space.

Varignon's Theorem: The Midpoint Parallelogram

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 any of $A$ , $B$ , $C$ , $D$ into any convex shape; the midpoints follow, and the readout under the canvas keeps the opposite sides of $PQRS$ equal. Learn more about why it holds
Labels toggles the point names. The Step explanations log keeps every step so far. Learn more about reading a scene

Proof?P, Q, R, S are the midpoints of the sides of ABCD. The diagonal AC makes two triangles, ABC and ADC; PQ and SR are their midsegments, so both are parallel to AC. The diagonal BD does the same for QR and PS. Both pairs of opposite sides of PQRS are parallel: it is a parallelogram, whatever shape ABCD has. Drag any vertex.Varignon’s theorem

DisplayLabelsMeasures

Drag A, B, C or D; the midpoints follow

PQ = 2.96, SR = 2.96;  QR = 2.35, PS = 2.35

Step 1 / 6

Step explanations

1The midpoints of the four sides

Given
ABCD is any quadrilateral. P, Q, R, S are the midpoints of AB, BC, CD, DA (one to four ticks). To prove: PQRS is a parallelogram.

The midpoints are computed; drag any vertex and they follow. Learn more about the statement

Getting Started with the Visualizer

The Statement

The First Diagonal

The First Midsegment

The Second Midsegment

Getting Started with the Visualizer

DemoA run through the proof player

Step 0 of 7

Step through the proof, then drag the four vertices into any shape you like and watch the inner figure stay a parallelogram. 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

3Draggable points — all four vertices A, B, C, D move freely; the hollow midpoints P, Q, R, S follow

4Readout — the line under the figure: the four sides of PQRS in opposite pairs, in the frame's units, 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 6 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 6 of 6

Proof?Varignon’s theoremDisplayLabelsMeasuresDrag A, B, C or D; the midpoints follow

PQ = 2.96, SR = 2.96;  QR = 2.35, PS = 2.35

← Back↻ ReplayNext →ResetNormal ▾Step 6 / 6

Step explanations

1The midpoints of the four sides

Given
ABCD is any quadrilateral. P, Q, R, S are the midpoints of AB, BC, CD, DA (one to four ticks). To prove: PQRS is a parallelogram.

2The diagonal AC

Construction
Draw the diagonal AC. It cuts ABCD into two triangles, ABC and ADC, and each of them has two of the midpoints on its sides.

3PQ ∥ AC in triangle ABC

Midsegment theorem
In triangle ABC, P is the midpoint of AB and Q the midpoint of BC, so PQ is a midsegment: PQ ∥ AC, and PQ = ½ AC.

4SR ∥ AC in triangle ADC

Midsegment theorem
In triangle ADC, S is the midpoint of DA and R the midpoint of CD, so SR is a midsegment: SR ∥ AC, and SR = ½ AC. Two lines parallel to AC are parallel to each other: PQ ∥ SR.

5QR ∥ BD and PS ∥ BD

Midsegment theorem
Now the other diagonal, BD. In triangle BCD, QR is a midsegment, so QR ∥ BD; in triangle BAD, PS is a midsegment, so PS ∥ BD. Hence QR ∥ PS.

6PQRS is a parallelogram

Definition
Both pairs of opposite sides of PQRS are parallel, PQ ∥ SR and QR ∥ PS, which is what a parallelogram is. Its sides are half the diagonals of ABCD: PQ = SR = ½ AC and QR = PS = ½ BD. ∎

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1The inner parallelogram — PQRS, lit at the last step; its sides are half the diagonals of ABCD

2The diagonals — AC and BD, dashed; each is parallel to one pair of sides of PQRS

3The chevrons — one for everything parallel to AC, two for everything parallel to BD

4Justification tags — Given, Construction, Midsegment theorem (three times), Definition

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

The Statement

Take any quadrilateral

ABCD

, not a parallelogram, not a trapezoid, nothing special, and mark the midpoints

P

Q

R

S

of its four sides. Join them in order. The theorem, published by Pierre Varignon in 1731, says that the figure

PQRS

is always a parallelogram, and more precisely that

PQ \parallel SR \parallel AC, \quad QR \parallel PS \parallel BD, \quad PQ = SR = \tfrac{1}{2} AC, \quad QR = PS = \tfrac{1}{2} BD

its sides are parallel to the diagonals of

ABCD

and half as long. The surprise is that the outer figure can be as lopsided as you like while the inner one keeps its symmetry.

The proof is the midsegment theorem applied four times, once in each of the four triangles that the two diagonals cut off. In the player the midpoints are hollow, computed from the vertices; all four vertices can be dragged. The readout under the canvas shows the four sides of

PQRS

in pairs, in the frame's own units, live while you drag.

Opening scene, frozen

ABCD with the midpoints P, Q, R, S of its sides joined into PQRS.

The First Diagonal

The construction draws the diagonal

AC

, dashed in the player. It cuts

ABCD

into two triangles,

ABC

below it and

ADC

above it, and it is the third side of each: in triangle

ABC

the midpoints

P

and

Q

sit on the other two sides, and in triangle

ADC

the midpoints

S

and

R

do.

That is the whole plan: each side of

PQRS

is a midsegment of one of these triangles, and the midsegment theorem says what a midsegment does. The diagonal

AC

takes care of two opposite sides of

PQRS

; the other diagonal,

BD

, will take care of the other two.

The first diagonal, frozen

AC dashed: two triangles, ABC and ADC.

The First Midsegment

In triangle

ABC

P

is the midpoint of

AB

and

Q

the midpoint of

BC

, so

PQ

joins the midpoints of two sides: it is a midsegment, and the midsegment theorem gives

PQ \parallel AC, \qquad PQ = \tfrac{1}{2} AC

The player fills the triangle and marks

PQ

and

AC

with one chevron each, pointing the same way. Nothing about the shape of

ABCD

was used; only that

P

and

Q

are midpoints.

The first midsegment, frozen

PQ in triangle ABC: parallel to AC, one chevron each.

The Second Midsegment

In triangle

ADC

, on the other side of the diagonal,

S

is the midpoint of

DA

and

R

the midpoint of

CD

, so

SR

is a midsegment of that triangle, and the same theorem gives

SR \parallel AC, \qquad SR = \tfrac{1}{2} AC

Now

PQ

and

SR

are both parallel to

AC

, and two lines parallel to the same line are parallel to each other, so

PQ \parallel SR

; and both are half of

AC

, so

PQ = SR

. One pair of opposite sides of

PQRS

is settled: equal and parallel.

The second midsegment, frozen

SR in triangle ADC: also parallel to AC, so PQ ∥ SR.

The Second Diagonal

The other diagonal,

BD

, does the same for the other pair of sides. It cuts

ABCD

into the triangles

BCD

and

BAD

. In triangle

BCD

Q

and

R

are the midpoints of

BC

and

CD

, so

QR

is a midsegment and

QR \parallel BD

QR = \tfrac{1}{2} BD

. In triangle

BAD

P

and

S

are the midpoints of

BA

and

AD

, so

PS

is a midsegment and

PS \parallel BD

PS = \tfrac{1}{2} BD

.

So

QR \parallel PS

and

QR = PS

: the second pair of opposite sides is equal and parallel too. The player marks the three parallel segments with two chevrons each.

The second diagonal, frozen

BD dashed; QR and PS are midsegments parallel to it, two chevrons each.

Why It Holds

A parallelogram is a quadrilateral whose opposite sides are parallel.

PQRS

has

PQ \parallel SR

and

QR \parallel PS

, so it is a parallelogram by definition; the opposite sides page then says its opposite sides are equal, which the midsegment theorem had already given directly. The end-of-proof mark ∎ in the log closes the argument. The tags name the facts used: the given midpoints, a construction, and the midsegment theorem three times over.

The proof never asked whether

ABCD

was convex, or even whether its sides cross. The midsegment theorem holds in any triangle, and the two diagonals always cut off four triangles, so the result is the same for a dart-shaped quadrilateral and for a crossed one. It even holds for four points that do not lie in one plane, since each step happens inside a single triangle.

The conclusion, frozen

Both pairs of opposite sides parallel: PQRS is a parallelogram.

Dragging the Figure

Drag any of

A

B

C

D

: the outer quadrilateral takes any shape, the four midpoints slide along their sides, and the readout keeps

PQ = SR

and

QR = PS

to the last digit. Pull

C

in and out and watch

PQ

and

SR

turn together, always parallel to the diagonal

AC

.

Make the diagonals equal and

PQRS

becomes a rhombus; make them perpendicular and it becomes a rectangle; both at once and it is a square. A drag that would make

ABCD

concave, or carry a vertex out of the frame, is refused; that is a limit of the drawing, not of the theorem.

After dragging C, frozen

C pulled up and in: the outer shape is very different, the inner figure still a parallelogram.

Consequences

• Half the area. The Varignon parallelogram has exactly half the area of the quadrilateral, since each corner triangle cut off, such as

APS

, is a quarter of the triangle on that diagonal.
• Perimeter. Its perimeter is

AC + BD

, the sum of the diagonals, since each side is half a diagonal.
• Special cases. If the diagonals of

ABCD

are perpendicular,

PQRS

is a rectangle; if they are equal, a rhombus; a kite therefore has a rectangle inside it and an isosceles trapezoid a rhombus.
• The bimedians. The diagonals of

PQRS

join the midpoints of opposite sides of

ABCD

; they are called the bimedians, and since the diagonals of a parallelogram bisect each other, the two bimedians of any quadrilateral bisect each other, at the centroid of the four vertices.
• Repeating. Taking the midpoint parallelogram of the midpoint parallelogram gives a figure similar to

ABCD

, shrunk by half and turned through a half-turn.