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Angles of a Parallelogram

How to use
  1. ▶ 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
  2. Drag AA, BB or DD (CC follows); the readout under the canvas shows the four angles and the sum of two neighbours. Learn more about why it holds
  3. 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?Each side of the parallelogram crosses the other pair of parallel sides, so the two angles at its ends are co-interior and add up to 180°: α + β = 180° along AB, β + γ = 180° along BC, γ + δ = 180° along CD. Subtracting gives α = γ and β = δ: opposite angles are equal. Drag A, B or D.Parallelogram angles
Display
Drag A, B or D (C keeps the figure a parallelogram)
ABDC
α = 63°, β = 117°, γ = 63°, δ = 117°, α + β = 180°
Step 1 / 6

Step explanations

1A parallelogram and its four angles
Given
ABCD is a parallelogram: AB ∥ DC (one chevron) and AD ∥ BC (two chevrons). Its angles are α at A, β at B, γ at C and δ at D. To prove: neighbouring angles add up to 180° and opposite angles are equal, α = γ and β = δ.
Drag a vertex: C keeps the figure a parallelogram. Learn more about the statement

Getting Started with the Visualizer
The Statement
The First Pair of Neighbours
The Second Pair of Neighbours
The First Pair of Opposites
The Third Pair of Neighbours
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 AA, BB or DD (CC follows so the figure stays a parallelogram) and watch the four angles in the readout: opposite ones stay equal and neighbours keep adding up to 180°180°. 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 D move freely; C follows so that both pairs of opposite sides stay parallel
4Readout — the line under the figure: the four angles and the sum of two neighbours, 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
12345
1The parallel sides — chevrons: one pair on AB and DC, two on AD and BC
2The angles at A and C — α and γ, equal at the end
3The angles at B and D — β and δ, equal at the end
4Justification tags — Given, Co-interior angles (three times), Subtracting, Combining
5Replay — at the last step Play becomes Replay and Next is off

The Statement

In a parallelogram ABCDABCD, with AB∥DCAB \parallel DC and AD∥BCAD \parallel BC, call the angles α\alpha at AA, β\beta at BB, γ\gamma at CC and δ\delta at DD. The theorem has two parts:

α+β=β+γ=γ+δ=δ+α=180°,α=γ,β=δ\alpha + \beta = \beta + \gamma = \gamma + \delta = \delta + \alpha = 180°, \qquad \alpha = \gamma, \quad \beta = \delta


neighbouring angles are supplementary, and opposite angles are equal. It completes the picture of the opposite sides and the diagonals pages: a parallelogram has equal opposite sides, equal opposite angles, and diagonals that bisect each other.

The whole proof is one fact used three times, the co-interior angles of the parallel lines page. In the player AA, BB, DD are free and CC is placed so that the figure stays a parallelogram; the readout shows the four angles and the sum of two neighbours.
ABDC
Opening scene, frozen

The parallelogram with both pairs of parallel sides marked.


The First Pair of Neighbours

The side ABAB crosses the two parallel sides ADAD and BCBC; it is a transversal of them. The angles α\alpha at AA and β\beta at BB both lie between the parallels and on the same side of ABAB: they are co-interior angles (also called same-side interior angles). On parallel lines co-interior angles add up to 180°180°:

α+β=180°\alpha + \beta = 180°
αβABDC
The first pair of neighbours, frozen

α at A and β at B, co-interior along AB: α + β = 180°.


The Second Pair of Neighbours

Now take BCBC as the transversal, crossing the other pair of parallels, ABAB and DCDC. The angles β\beta at BB and γ\gamma at CC are co-interior between them:

β+γ=180°\beta + \gamma = 180°
αβγABDC
The second pair of neighbours, frozen

β and γ, co-interior along BC: β + γ = 180°.


The First Pair of Opposites

The angles α\alpha and γ\gamma both make 180°180° with the same angle β\beta:

α=180°−β=γ\alpha = 180° - \beta = \gamma


so the opposite angles at AA and CC are equal.
αβγABDC
The first pair of opposites, frozen

α and γ lit: both are 180° − β.


The Third Pair of Neighbours

The side CDCD crosses ADAD and BCBC, so γ\gamma and δ\delta are co-interior: γ+δ=180°\gamma + \delta = 180°. Then

δ=180°−γ=β\delta = 180° - \gamma = \beta


and the opposite angles at BB and DD are equal too. The fourth pair, δ+α=180°\delta + \alpha = 180°, follows from the others, or from DADA as a transversal.
αβγδABDC
The third pair of neighbours, frozen

γ + δ = 180° along CD, so δ = β.


Why It Holds

Every side of a parallelogram is a transversal of the other two sides, which are parallel, so the angles at its two ends are co-interior and add up to 180°180°. Going round the figure, each angle is the supplement of both its neighbours, and two angles with the same supplement are equal:

α=γ,β=δ,α+β=180°\alpha = \gamma, \qquad \beta = \delta, \qquad \alpha + \beta = 180°


The end-of-proof mark ∎ in the log closes the argument. The tags name the facts used: the given figure, co-interior angles three times, a subtraction, and combining. As a check, the four angles add up to 2(α+β)=360°2(\alpha + \beta) = 360°, the angle sum of a quadrilateral.
αβγδABDC
The conclusion, frozen

Opposite angles equal, neighbours supplementary.


Dragging the Figure

Drag AA, BB or DD: the parallelogram leans and stretches, CC moves so that both pairs of opposite sides stay parallel, and the readout shows the angles at AA and CC staying equal, the angles at BB and DD staying equal, and α+β\alpha + \beta staying at 180°180°. Make the figure a rectangle and all four angles become 90°90°.

The player keeps the parallelogram convex, with sides long enough for the angle marks and every angle between 25°25° and 155°155°; a drag that would flatten it or carry a vertex out of the frame is refused.
αβγδABDC
After dragging D, frozen

D moved to the right: sharper angles at A and C, the opposite angles still equal.


Consequences

• Rectangles. If one angle of a parallelogram is right, all four are: its neighbours are 180°−90°180° - 90° and its opposite equals it. That is why a parallelogram with one right angle is a rectangle, the starting point of the rectangle diagonals page.
• The converse. A quadrilateral with both pairs of opposite angles equal is a parallelogram: the angles add up to 360°360°, so α+β=180°\alpha + \beta = 180°, and equal co-interior sums make the sides parallel.
• Angle bisectors. The bisectors of two neighbouring angles meet at a right angle, since half of 180°180° is 90°90°, as for the bisectors of a linear pair; the four bisectors of a parallelogram enclose a rectangle.
• Rhombus and kite. In a rhombus each diagonal bisects the angles at its ends, and the equal opposite angles are split into equal halves; the rhombus diagonals page shows the diagonals are perpendicular.
• Cyclic parallelograms. A parallelogram is cyclic only if its opposite angles add up to 180°180°; equal and supplementary means 90°90° each, so the only cyclic parallelograms are rectangles.