▶ 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 D; the fourth vertex C follows so the figure stays a parallelogram, and the readout under the canvas keeps the opposite sides equal. 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?Draw the diagonal BD. It is a transversal of the parallels AB, DC and also of AD, BC, so it makes two pairs of equal alternate angles, α at B and D, β at D and B. Triangles ABD and CDB then share the side BD between two equal angles: they are congruent (ASA), so AB = CD and AD = BC. Drag A, B or D; C follows to keep the figure a parallelogram.Parallelogram: opposite sides
Display
Drag A, B or D; C follows so the figure stays a parallelogram
AB = 4.00, CD = 4.00; AD = 3.32, BC = 3.32
Step 1 / 6
Step explanations
1A parallelogram
Given ABCD has AB ∥ DC (one chevron each) and AD ∥ BC (two chevrons each): it is a parallelogram. To prove: AB = CD and AD = BC, opposite sides are equal.
Step through the proof, then drag the vertices and watch the fourth one follow so that the figure stays a parallelogram and the opposite sides 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
Proof?Parallelogram: opposite sidesDisplayLabelsMeasuresDrag A, B or D; C follows so the figure stays a parallelogram
AB = 4.00, CD = 4.00; AD = 3.32, BC = 3.32
← Back▶ PlayNext →ResetNormal ▾Step 1 / 6
Step explanations
1A parallelogram
Given ABCD has AB ∥ DC (one chevron each) and AD ∥ BC (two chevrons each): it is a parallelogram. To prove: AB = CD and AD = BC, opposite sides 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, Measures writes each marked angle's size beside its arc
3Draggable points — A, B and D move freely; the hollow fourth vertex C follows to keep the figure a parallelogram
4Readout — the line under the figure: the four side lengths 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?Parallelogram: opposite sidesDisplayLabelsMeasuresDrag A, B or D; C follows so the figure stays a parallelogram
AB = 4.00, CD = 4.00; AD = 3.32, BC = 3.32
← Back↻ ReplayNext →ResetNormal ▾Step 6 / 6
Step explanations
1A parallelogram
Given ABCD has AB ∥ DC (one chevron each) and AD ∥ BC (two chevrons each): it is a parallelogram. To prove: AB = CD and AD = BC, opposite sides are equal.
2The diagonal BD
Construction Join B to D. The diagonal cuts the parallelogram into two triangles, ABD and CDB, and it is a transversal of both pairs of parallel sides.
3∠ABD = ∠CDB = α
Alternate angles AB ∥ DC and BD crosses both. The angles ∠ABD and ∠CDB lie on opposite sides of BD, between the parallels: they are alternate angles, so they are equal.
4∠ADB = ∠CBD = β
Alternate angles AD ∥ BC and BD crosses both. The angles ∠ADB and ∠CBD are alternate angles on this second pair of parallels, so they are equal too.
5Triangles ABD and CDB are congruent
ASA congruence In triangle ABD the side BD lies between the angles α at B and β at D; in triangle CDB the same side BD lies between α at D and β at B. Two angles and the included side: the triangles are congruent, with A ↔ C, B ↔ D, D ↔ B.
6AB = CD, AD = BC
Corresponding parts Corresponding sides of congruent triangles are equal: AB ↔ CD and AD ↔ CB. The corresponding angles at A and at C are equal as well, and ∠B = ∠D = α + β: opposite angles of a parallelogram are equal. ∎
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1The diagonal BD — the common side of the two triangles, and a transversal of both pairs of parallels
2Alternate angles — α at B and D on the parallels AB, DC; β at D and B on the parallels AD, BC
3The chevrons — one for the pair AB ∥ DC, two for the pair AD ∥ BC
5Replay — at the last step Play becomes Replay and Next is off
The Statement
A parallelogram is a quadrilateral whose opposite sides are parallel: in ABCD, AB∥DC and AD∥BC. That is the whole definition; nothing is said about lengths or angles. The theorem, Proposition 34 of Book I of Euclid's Elements, says that the lengths and angles follow:
AB=CD,AD=BC,∠A=∠C,∠B=∠D
opposite sides are equal and opposite angles are equal. Euclid also adds that a diagonal bisects the area, which is the same picture.
The proof is the first on this section's quadrilateral pages and sets the pattern for the others: draw a diagonal, find two congruent triangles, read off the equal parts. The congruence comes from the parallel lines page, since a diagonal is a transversal of both pairs of parallel sides. In the player C is hollow: it is computed as the fourth vertex that makes ABCD a parallelogram, so dragging any of A, B, D moves it too. The readout under the canvas shows the four side lengths in the frame's own units, live while you drag.
Opening scene, frozen
ABCD with its two pairs of parallel sides marked by chevrons; C hollow, built from A, B, D.
The Diagonal
The construction is one segment: the diagonal BD. It cuts the parallelogram into two triangles, ABD in one colour and CDB in the other, which share the side BD. The proof will show that these two triangles are congruent; everything else is reading off corresponding parts.
The diagonal does a second job. It crosses the parallel sides AB and DC, and it also crosses the parallel sides AD and BC: it is a transversal of both pairs of parallels at once. That gives two pairs of alternate angles, one at each end of the diagonal, which is exactly what a congruence by angles needs.
The diagonal, frozen
BD cuts the parallelogram into the triangles ABD and CDB.
Alternate Angles Once
Take the first pair of parallels, AB∥DC, with the transversal BD. The angle ∠ABD at B and the angle ∠CDB at D lie on opposite sides of the transversal and between the two parallels: they are alternate angles. By the parallel lines theorem, alternate angles on parallel lines are equal:
∠ABD=∠CDB=α
The player draws both in the same colour. One is an angle of triangle ABD at B; the other is an angle of triangle CDB at D.
Alternate angles once, frozen
On the parallels AB, DC: ∠ABD = ∠CDB = α.
Alternate Angles Twice
Now the second pair of parallels, AD∥BC, with the same transversal BD. The angles ∠ADB at D and ∠CBD at B are alternate angles on this pair, so they are equal as well:
∠ADB=∠CBD=β
At each end of the diagonal there are now two marked angles, α and β, and together they make up the angle of the parallelogram there: ∠B=α+β and ∠D=β+α. That already shows that the angles at B and at D are equal.
Alternate angles twice, frozen
On the parallels AD, BC: ∠ADB = ∠CBD = β.
Two Congruent Triangles
Compare the two triangles along their common side. In triangle ABD the side BD lies between the angle α at B and the angle β at D. In triangle CDB the same side BD lies between the angle α at D and the angle β at B. Two angles and the side between them are equal, so the triangles are congruent by angle-side-angle, with the vertices matched A↔C, B↔D, D↔B.
Why angle-side-angle is enough: once two angles of a triangle are fixed, the third is fixed by the angle sum, and a triangle with a given side and the two angles at its ends can be drawn in only one way. The matching of vertices matters, because it tells which side corresponds to which.
The congruent triangles, frozen
BD between α and β in both triangles: congruent by ASA.
Why It Holds
Corresponding parts of congruent triangles are equal. With A↔C, B↔D, D↔B, the side AB corresponds to CD and the side AD corresponds to CB, so AB=CD and AD=BC: opposite sides are equal. The angle at A corresponds to the angle at C, so ∠A=∠C; and ∠B=∠D=α+β was seen already: opposite angles are equal. The end-of-proof mark ∎ in the log closes the argument. The tags name the facts used: the given parallels, a construction, alternate angles twice, angle-side-angle congruence, and corresponding parts.
The same two triangles give Euclid's last clause, that the diagonal halves the parallelogram: congruent triangles have equal areas. The other diagonal, AC, does the same with triangles ABC and CDA, which is what the next quadrilateral page, on the diagonals bisecting each other, starts from.
The conclusion, frozen
AB = CD and AD = BC, the four sides lit.
Dragging the Figure
Drag D sideways: the parallelogram shears, the angles α and β change, the diagonal tilts, and the readout keeps AB=CD and AD=BC to the last digit. Drag D straight up or down and the figure gets taller or flatter. Drag A or B to change the base; C follows in every case, since it is built as D plus the vector from A to B.
Make D sit straight above A and the parallelogram becomes a rectangle, with α+β=90° at every vertex. A drag that would flatten the figure, putting D on the line AB, is refused, as is one that would carry a vertex out of the frame.
After dragging D, frozen
D pushed right: a sheared, flatter parallelogram; the readout still pairs the sides.
Consequences
• Opposite angles.∠A=∠C and ∠B=∠D, from the same congruence; and neighbouring angles add up to 180°, since all four add up to 360° by the polygon angle sum. • The converse. A quadrilateral with both pairs of opposite sides equal is a parallelogram: the diagonal gives two triangles congruent by side-side-side, and the equal alternate angles make the sides parallel. • Area. A diagonal halves a parallelogram, so its area is twice a triangle's: base times height. • Special cases. A rectangle, a rhombus and a square are parallelograms, so each has equal opposite sides; a rhombus is the case where all four sides are equal. A parallelogram inscribed in a circle must be a rectangle, by the cyclic quadrilateral theorem, since its equal opposite angles must add up to 180°. • Vectors.C=D+(B−A) is the parallelogram rule for adding vectors: the sum of AB and AD is the diagonal AC.