How long is the diagonal of a square?

The side times √2. In the right triangle formed by two sides and a diagonal, the Pythagorean theorem gives d² = s² + s² = 2s², so d = s√2, about 1.414 times the side.

Are the diagonals of a square perpendicular?

Yes. Each diagonal meets the sides at 45°, so in each of the four triangles around the crossing point the two angles at the square's corners are 45° and the angle at the centre is 90°.

What angle does the diagonal of a square make with a side?

45°. The diagonal cuts the square into two isosceles right triangles, whose base angles are (180° − 90°)/2 = 45°.

Diagonals of a Square

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 $A$ or $B$ to turn and resize the square; the readout under the canvas shows both diagonals, the side times $\sqrt2$ and the angle at $O$ . 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?The diagonal AC cuts the square into two isosceles right triangles, so it meets the sides at 45°. The triangles ABC and BAD are congruent by SAS, so AC = BD. In the triangle OAB the angles at A and B are both 45°, so the angle at O is 90° and OA = OB; the same around O gives OA = OB = OC = OD. By Pythagoras AC = AB · √2. Drag A or B.Square diagonals

DisplayLabelsMeasures

Drag A or B to turn and resize the square

AC = 4.71,  BD = 4.71,  AB·√2 = 4.71,  ∠AOB = 90°

Step 1 / 6

Step explanations

1A square

Given
ABCD is a square: four equal sides (one tick each) and four right angles. To prove: its diagonals AC and BD are equal, perpendicular, bisect each other, make 45° with the sides, and are √2 times the side.

Drag A or B: C and D keep the square. Learn more about the statement

Getting Started with the Visualizer

The Statement

Half a Square

Equal Diagonals

A Right Angle at the Centre

The Diagonals Bisect Each Other

Why It Holds

Dragging the Figure

Consequences

Getting Started with the Visualizer

DemoA run through the proof player

Step 0 of 7

Step through the proof, then drag

A

B

to turn and resize the square, and watch the readout: the two diagonals stay equal to each other and to the side times

\sqrt2

, and they cross at

90°

. 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

Proof?Square diagonalsDisplayLabelsMeasuresDrag A or B to turn and resize the square

AC = 4.71,  BD = 4.71,  AB·√2 = 4.71,  ∠AOB = 90°

← Back▶ PlayNext →ResetNormal ▾Step 1 / 6

Step explanations

1A square

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 and B move freely and turn or resize the square; the hollow points C, D and O follow

4Readout — the line under the figure: AC, BD, AB·√2 and the angle at O, 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?Square diagonalsDisplayLabelsMeasuresDrag A or B to turn and resize the square

AC = 4.71,  BD = 4.71,  AB·√2 = 4.71,  ∠AOB = 90°

← Back↻ ReplayNext →ResetNormal ▾Step 6 / 6

Step explanations

1A square

2The diagonal meets the sides at 45°

Isosceles base angles
The diagonal AC cuts off the triangle ABC with AB = BC and a right angle at B. It is isosceles, so its base angles at A and C are equal, φ each, and the angle sum gives 2φ + 90° = 180°: φ = 45°. The diagonal halves the right angles at A and C.

3AC = BD

SAS
Draw the other diagonal BD. The triangles ABC and BAD have the common side AB, BC = AD, and right angles at B and A between them. By SAS they are congruent, so AC = BD. The same argument as before gives ∠ABD = 45°.

4The diagonals meet at 90°

Angle sum
The diagonals cross at O. In the triangle OAB the angles at A and B are both 45° (∠OAB = ∠CAB and ∠OBA = ∠DBA), so the angle at O is 180° − 45° − 45° = 90°: the diagonals are perpendicular.

5OA = OB = OC = OD

Equal angles, equal sides
The triangle OAB has equal angles at A and B, so the sides facing them are equal: OA = OB. Every one of the four triangles around O has two 45° angles in the same way, so OA = OB = OC = OD (two ticks each): the diagonals bisect each other.

6AC = AB · √2

Pythagorean theorem
In the right triangle ABC, AC² = AB² + BC² = 2 · AB², so AC = AB · √2, and BD too. The diagonals of a square are equal, perpendicular, bisect each other, meet the sides at 45° and are √2 times the side. ∎

12345

1The two diagonals — AC and BD, equal and crossing at right angles at O

2The 45° angles — φ at A, B and C, where the diagonals halve the right angles

3The half diagonals — OA, OB, OC, OD, two ticks each

4Justification tags — Given, Isosceles base angles, SAS, Angle sum, Equal angles equal sides, Pythagorean theorem

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

The Statement

A square has four equal sides and four right angles. Its diagonals

AC

and

BD

have every good property a quadrilateral's diagonals can have:

AC = BD = AB\sqrt2, \qquad AC \perp BD, \qquad OA = OB = OC = OD

and each diagonal meets the sides at

45°

, halving the right angles. A square is both a rectangle and a rhombus, so these facts also follow from the rectangle diagonals (equal) and the rhombus diagonals (perpendicular) pages; this page proves them directly, from one observation: a diagonal cuts the square into two isosceles right triangles.

In the player

A

and

B

are free, and

D

and

C

are built so that

ABCD

is always a square; the readout shows both diagonals, the side times

\sqrt2

and the angle between the diagonals.

Opening scene, frozen

The square ABCD: four equal sides, four right angles.

Half a Square

The diagonal

AC

cuts off the triangle

ABC

, with

AB = BC

(sides of the square) and a right angle at

B

. It is isosceles, so its base angles at

A

and

C

are equal,

\varphi

each, and the angle sum gives

2\varphi + 90° = 180°, \qquad \varphi = 45°

The diagonal halves the right angles at

A

and

C

, meeting the sides at

45°

Half a square, frozen

ABC shaded: isosceles with a right angle at B, base angles φ = 45°.

Equal Diagonals

Draw the second diagonal

BD

. The triangles

ABC

and

BAD

share the side

AB

, have

BC = AD

, and have right angles at

B

and

A

between those sides. By SAS they are congruent, so

AC = BD

The same isosceles argument in

BAD

gives

\angle ABD = 45°

Equal diagonals, frozen

BD drawn; ABC and BAD congruent by SAS: AC = BD.

A Right Angle at the Centre

Let the diagonals cross at

O

. In the triangle

OAB

the angle at

A

\angle CAB = 45°

and the angle at

B

\angle DBA = 45°

, so by the angle sum

\angle AOB = 180° - 45° - 45° = 90°

The diagonals are perpendicular.

A right angle at the centre, frozen

OAB shaded: two 45° angles, so 90° at O.

The Diagonals Bisect Each Other

The triangle

OAB

has equal angles at

A

and

B

, so the sides facing them are equal:

OA = OB

. Each of the four triangles

OAB

OBC

OCD

ODA

around

O

has two

45°

angles in the same way, so

OA = OB = OC = OD

The diagonals cut each other in half, and

O

is equally far from all four corners: the centre of the square's circumcircle.

The diagonals bisect each other, frozen

OA = OB = OC = OD, two ticks each.

Why It Holds

Finally, the length. In the right triangle

ABC

the Pythagorean theorem gives

AC^2 = AB^2 + BC^2 = 2 \cdot AB^2, \qquad AC = AB\sqrt2

The end-of-proof mark ∎ in the log closes the argument. The tags name the facts used: the given figure, isosceles base angles, SAS, the angle sum, equal angles facing equal sides, and the Pythagorean theorem.

Everything comes from the four triangles with angles

45°

45°

90°

: the two halves of the square cut by a diagonal, and the four quarters cut by both.

The conclusion, frozen

AC = BD = AB · √2.

Dragging the Figure

Drag

A

B

: the square turns and grows or shrinks,

C

and

D

follow, and the readout shows

AC = BD = AB\sqrt2

and

\angle AOB = 90°

at every position. The ratio of diagonal to side never changes: it is always

\sqrt2 \approx 1.414

.

The player keeps the side long enough for the marks and the whole square inside the frame; a drag that would push a corner out is refused.

After dragging B, frozen

B moved up a little: the square turned, its diagonals still equal and perpendicular.

Consequences

• The diagonal formula. A square of side

s

has diagonal

s\sqrt2

; conversely a square with diagonal

d

has side

d/\sqrt2

and area

d^2/2

, half the square on the diagonal.
• Doubling the square. The square on the diagonal has twice the area of the original, the construction Socrates draws in Plato's Meno.
• √2 is irrational. The diagonal and the side of a square have no common measure; this was the first irrational ratio found, by the Pythagoreans.
• The 45-45-90 triangle. Half a square is the right isosceles triangle with sides in the ratio

1 : 1 : \sqrt2

, which gives

\sin 45° = \cos 45° = 1/\sqrt2

.
• Rectangle and rhombus. A square is both: equal diagonals from the rectangle page, perpendicular ones from the rhombus page, both bisecting from the parallelogram page. A quadrilateral whose diagonals are equal, perpendicular and bisect each other is a square.