▶ 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 up or down for the height, or A and B for the base; C follows, and the readout under the canvas keeps the two diagonals equal. Learn more about why it holds
Proof?A rectangle has four right angles, so its opposite sides are parallel and it is a parallelogram: AD = BC. The triangles DAB and CBA share the base AB, have AD = BC, and have right angles between those sides, so they are congruent (SAS) and their third sides are equal: DB = CA. Drag D up or down to change the height, A or B to change the base.Rectangle: diagonals
Display
Drag D up or down for the height, A or B for the base
AC = 5.17, BD = 5.17
Step 1 / 6
Step explanations
1A rectangle and its diagonals
Given ABCD has four right angles: it is a rectangle. Its diagonals are AC and BD. To prove: AC = BD, the diagonals are equal.
Step through the proof, then change the height and the base of the rectangle and watch its two diagonals stay the same 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
Proof?Rectangle: diagonalsDisplayLabelsMeasuresDrag D up or down for the height, A or B for the base
AC = 5.17, BD = 5.17
← Back▶ PlayNext →ResetNormal ▾Step 1 / 6
Step explanations
1A rectangle and its diagonals
Given ABCD has four right angles: it is a rectangle. Its diagonals are AC and BD. To prove: AC = BD, the diagonals are equal.
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1Proof — the theorem loaded in the player; the ? gives its idea in one line
2Display — Labels shows the point names
3Draggable points — D slides up and down the perpendicular at A, which keeps the angles right; A and B move the base; the hollow C follows
4Readout — the line under the figure: the two diagonals 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?Rectangle: diagonalsDisplayLabelsMeasuresDrag D up or down for the height, A or B for the base
AC = 5.17, BD = 5.17
← Back↻ ReplayNext →ResetNormal ▾Step 6 / 6
Step explanations
1A rectangle and its diagonals
Given ABCD has four right angles: it is a rectangle. Its diagonals are AC and BD. To prove: AC = BD, the diagonals are equal.
2A rectangle is a parallelogram
Co-interior angles The angles at A and D are co-interior angles on the transversal AD of the lines AB and DC, and they add up to 90° + 90° = 180°, so AB ∥ DC. In the same way AD ∥ BC. A rectangle is a parallelogram.
3AD = BC
Opposite sides Opposite sides of a parallelogram are equal, so AD = BC (one tick each). These are the two sides that stand on the base AB at its ends.
4Two triangles on the base AB
Construction The diagonal BD cuts off the triangle DAB, and the diagonal AC cuts off the triangle CBA. Both stand on the base AB; one has its right angle at A, the other at B.
5Triangles DAB and CBA are congruent
SAS congruence AD = BC, the base AB is shared, and the angles between those sides, at A in one triangle and at B in the other, are both right angles. Two sides and the included angle: the triangles are congruent, with D ↔ C, A ↔ B, B ↔ A.
6AC = BD
Corresponding parts Corresponding sides of congruent triangles are equal: DB ↔ CA. So the two diagonals have the same length (two ticks each). ∎
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1The equal diagonals — AC and BD, two ticks each at the last step
2The two triangles — DAB and CBA on the base AB, overlapping in the middle; congruent by SAS
3The right angles — at A and at B, the included angles of the two triangles
4Justification tags — Given, Co-interior angles, Opposite sides, Construction, SAS congruence, Corresponding parts
5Replay — at the last step Play becomes Replay and Next is off
The Statement
A rectangle is a quadrilateral with four right angles. Its diagonals AC and BD are, in general, not perpendicular and not the axes of any symmetry, but they are equal:
AC=BD
This is the property that singles out rectangles among parallelograms, just as perpendicular diagonals single out rhombi; a square has both. The proof is three short steps: a rectangle is a parallelogram, so its opposite sides are equal; the two diagonals cut off two right triangles on the same base; and those triangles are congruent by side-angle-side.
In the player D slides up and down on the perpendicular to AB at A, which is what keeps the angle at A right, and C is computed from it, so the figure is always a rectangle. The readout under the canvas shows both diagonals in the frame's own units, live while you drag.
Opening scene, frozen
ABCD with four right-angle marks and its two diagonals.
A Rectangle Is a Parallelogram
Nothing in the definition of a rectangle says that its opposite sides are parallel, so that has to be shown. Take the side AD as a transversal of the lines AB and DC. The angles at A and at D lie on the same side of the transversal, between the two lines: they are co-interior angles, and they add up to 90°+90°=180°. Co-interior angles that are supplementary mean the lines are parallel, by the converse of the parallel lines theorem, so AB∥DC. The same argument with AB as the transversal gives AD∥BC.
Both pairs of opposite sides are parallel: a rectangle is a parallelogram, and the player marks the two pairs with one and two chevrons. Everything proved about parallelograms now applies to it.
A parallelogram, frozen
Co-interior angles of 90° + 90°: both pairs of opposite sides are parallel (chevrons).
the two sides that stand on the base AB at its ends, marked with one tick each. It also gives AB=DC, which is not needed here. From now on the proof uses only the base AB, the two equal sides on it, and the right angles at A and B.
Opposite sides equal, frozen
AD = BC, one tick each.
Two Triangles on the Base
Each diagonal cuts the rectangle into two triangles; the proof takes one triangle from each cut, the two that stand on the base AB. The diagonal BD cuts off the triangle DAB, with its right angle at A; the diagonal AC cuts off the triangle CBA, with its right angle at B. The player fills them in two colours; they overlap on the middle of the rectangle, which is why the colours mix there.
The two triangles are mirror images of each other in the perpendicular bisector of AB. The proof does not use the mirror, but it is a good way to see in advance why the diagonals, which are the hypotenuses of these triangles, must be equal.
The two triangles, frozen
DAB with its right angle at A, CBA with its right angle at B, both on the base AB.
Side Angle Side
Compare the two triangles along the base. In triangle DAB the sides AD and AB meet at A in a right angle. In triangle CBA the sides BC and BA meet at B in a right angle. Now AD=BC, AB=BA is the same segment, and the included angles are both 90°. Two sides and the angle between them are equal, so the triangles are congruent by side-angle-side, with the vertices matched D↔C, A↔B, B↔A.
The matching swaps A and B, which is the mirror of the previous section written out: the triangle DAB read from D is the triangle CBA read from C.
Side angle side, frozen
AD = BC, AB shared, right angles between: congruent.
Why It Holds
Corresponding sides of congruent triangles are equal. With D↔C and B↔A, the side DB corresponds to CA, so BD=AC: the diagonals are equal. The end-of-proof mark ∎ in the log closes the argument. The tags name the facts used: the given right angles, co-interior angles, equal opposite sides, a construction, side-angle-side congruence, and corresponding parts.
The same proof, run backwards, gives the converse: in a parallelogram with equal diagonals the triangles DAB and CBA are congruent by side-side-side, so the angles at A and B are equal, and since they are co-interior angles on AB between the parallels AD and BC they add up to 180°; so each is 90° and the parallelogram is a rectangle.
The conclusion, frozen
AC = BD, two ticks each.
Dragging the Figure
Drag D up or down: the rectangle gets taller or flatter, both diagonals change length together, and the readout keeps AC=BD to the last digit. When the height equals the base the rectangle is a square and the diagonals are also perpendicular. Drag A or B to change the base or to turn the whole rectangle; D keeps its distance from A along the perpendicular, and C follows.
D cannot leave the perpendicular at A, since that would spoil the right angle. A drag that would make the rectangle too flat or too tall for the frame, or shrink the base to nothing, is refused.
After dragging D, frozen
D pulled down: a flatter rectangle, both diagonals shorter and still equal.
Consequences
• The converse. A parallelogram with equal diagonals is a rectangle, as shown under Why It Holds. Builders use it: a frame with equal diagonals is square at the corners. • A rectangle is cyclic. The diagonals also bisect each other, so their common midpoint is at the same distance from all four vertices: it is the centre of a circle through them, with each diagonal a diameter. This is the cyclic quadrilateral theorem seen from the other side, and the right angles are those of Thales' theorem. • The median to the hypotenuse. Half a rectangle is a right triangle, and half a diagonal is the median from the right angle to the hypotenuse; so that median is half the hypotenuse. • Length of the diagonal. By the Pythagorean theorem, a rectangle with sides a and b has diagonals of length a2+b2. • Squares. A square is a rectangle and a rhombus at once, so its diagonals are equal, perpendicular, and bisect each other.