Parallelograms

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Learning goals

Identify the parallel-side hypotheses in a parallelogram
Prove opposite-side and angle relationships using parallel lines and congruence
Calculate missing lengths and angles from parallelogram properties
Justify diagonal bisection without assuming a symmetry
Compare half-turn and reflection symmetry in general and special parallelograms

Start with the definition

A parallelogram is a quadrilateral with both pairs of opposite sides parallel. Throughout this lesson, vertices are named in boundary order and the quadrilaterals are convex, with four distinct corners.

For parallelogram $ABCD$ , the hypotheses are

\begin{gathered} AB\parallel CD,\\ BC\parallel AD. \end{gathered}

A tilted drawing supplies neither of these facts on its own. A rectangle, rhombus, and square are special parallelograms, so the properties proved here apply to them too. Under our inclusive convention, every parallelogram is also a trapezoid because a trapezoid needs at least one parallel opposite-side pair.

Alternate interior angles made by a transversal across parallel lines are congruent. ASA proves two triangles congruent from two corresponding angles and their included side; use Parallel Lines and Transversals or SAS and ASA Congruence for a refresher.

Check your understanding

Which information is the definition's starting point for a parallelogram $ABCD$ ?

A Both pairs of opposite sides are parallel B Both diagonals have the same length C Every angle is right D Every side has the same length

A diagonal reveals opposite sides and angles

Opposite sides of a parallelogram have matching lengths. For example, if two neighboring sides are $7$ cm and $4$ cm, the sides opposite them are $7$ cm and $4$ cm too. The reason is the matching triangles formed by a diagonal, rather than the apparent balance of the sketch.

The chevrons mark the two given parallel pairs. The diagonal is an auxiliary line.

Opposite sides and angles of a parallelogram#

Let $ABCD$ be any parallelogram and draw $AC$ . Using $AC$ as a transversal of the two given parallel pairs gives

\begin{gathered} \angle BAC\cong\angle DCA,\\ \angle BCA\cong\angle DAC. \end{gathered}

The included side $AC$ is common to both triangles. ASA therefore gives the ordered correspondence

\triangle ABC\cong\triangle CDA.

Corresponding sides and angles of these congruent triangles give

\begin{gathered} AB=CD,\\ BC=DA,\\ \angle ABC\cong\angle CDA. \end{gathered}

The other two whole angles consist of the two matched angle pieces at each end of $AC$ . Adding their measures gives

\begin{aligned} m\angle DAB&=m\angle DAC+m\angle CAB,\\ m\angle BCD&=m\angle BCA+m\angle ACD,\\ m\angle DAB&=m\angle BCD. \end{aligned}

No side length or particular angle size was needed: both opposite-side pairs and both opposite-angle pairs match in every parallelogram.

Worked example 1 Use the correct opposite side

Problem. $ABCD$ is a parallelogram. The lengths in centimeters satisfy

\begin{gathered} AB=3x+2,\\ CD=23,\\ BC=9. \end{gathered}

Find $x$ and the length of $AD$ .

Use the given parallelogram hypothesis to connect opposite sides.

Solution. Opposite sides $AB$ and $CD$ are congruent, so

\begin{aligned} 3x+2&=23,\\ 3x&=21,\\ x&=7. \end{aligned}

The side opposite $BC$ is $AD$ , so $AD$ is $9$ cm. The two neighboring sides need not match each other.

Check your understanding

In parallelogram $ABCD$ , diagonal $AC$ is drawn. Which reason proves that $\angle BAC$ and $\angle DCA$ are congruent before triangle congruence is known?

A They are vertical angles B They belong to an isosceles triangle C They are alternate interior angles for parallel sides D All angles of a parallelogram are congruent

Consecutive angles are angles at neighboring vertices. In a parallelogram they are supplementary, meaning their measures add to $180^\circ$ . For instance, a $72^\circ$ corner has $108^\circ$ neighbors: those are same-side interior angles across parallel lines.

In the diagram above, $AB$ is a transversal of the parallel lines $AD$ and $BC$ . The same-side interior-angle theorem gives

m\angle DAB+m\angle ABC=180^\circ.

Taking each side in turn as the transversal proves this for every consecutive pair. Together with opposite-angle congruence, one corner therefore determines the other three.

Worked example 2 Choose the angle relationship first

Problem. $ABCD$ is a parallelogram with

\begin{gathered} m\angle A=(2x+14)^\circ,\\ m\angle B=(5x+40)^\circ. \end{gathered}

Find $x$ and the measures of angles $A$ and $C$ .

Solution. Angles $A$ and $B$ are consecutive, so add their measures rather than equating them.

\begin{aligned} (2x+14)+(5x+40)&=180,\\ 7x+54&=180,\\ 7x&=126,\\ x&=18. \end{aligned}

Substitution gives

\begin{aligned} m\angle A&=2(18)+14,\\ m\angle A&=50^\circ. \end{aligned}

Angle $C$ is opposite $A$ , so its measure is also $50^\circ$ . Angle $B$ is $130^\circ$ , which checks the straight-angle total.

Worked example 3 A diagonal splits one corner

Problem. In parallelogram $ABCD$ , diagonal $AC$ lies inside angle $DAB$ . Its two parts measure

\begin{gathered} m\angle BAC=29^\circ,\\ m\angle CAD=38^\circ. \end{gathered}

Find the whole angles $DAB$ and $BCD$ , and then angle $ABC$ .

The diagonal separates the two given parts of the corner at A.

Solution. The diagonal divides the whole angle into two parts, which need not be equal.

\begin{aligned} m\angle DAB&=29^\circ+38^\circ,\\ m\angle DAB&=67^\circ. \end{aligned}

The opposite whole angle matches it.

m\angle BCD=67^\circ.

The consecutive angle is supplementary to it.

\begin{aligned} m\angle ABC&=180^\circ-67^\circ,\\ m\angle ABC&=113^\circ. \end{aligned}

Check your understanding

One angle of a parallelogram is $118^\circ$ . What is the measure of either neighboring angle?

118^\circ

62^\circ

59^\circ

242^\circ

Each diagonal cuts the other in half

To bisect a segment means to divide it into two equal lengths. If diagonals of a parallelogram meet at $E$ , a $14$ cm diagonal has two $7$ cm parts, while a $10$ cm diagonal has two $5$ cm parts. Bisection compares parts of the same diagonal; it does not make the two diagonals equal.

Compare the triangles at opposite ends of the diagonals; the crossing is not assumed to be a midpoint.

Diagonals of a parallelogram bisect each other#

Let the diagonals of parallelogram $ABCD$ intersect at $E$ . The intersection is inside the convex quadrilateral. We already proved that opposite sides have equal lengths, so

AB=CD.

Since $AB$ is parallel to $CD$ , the two diagonals give alternate interior-angle pairs:

\begin{gathered} \angle BAE\cong\angle DCE,\\ \angle ABE\cong\angle CDE. \end{gathered}

The equal sides lie between these equal angles, so ASA gives

\triangle ABE\cong\triangle CDE.

Corresponding parts now give

\begin{gathered} AE=CE,\\ BE=DE. \end{gathered}

Thus $E$ is the midpoint of each diagonal. This proof used parallel sides and the proved opposite-side property, without assuming a symmetry or assuming that the crossing was a midpoint.

Worked example 4 Keep the two diagonals separate

Problem. The diagonals of parallelogram $ABCD$ meet at $E$ . Lengths in centimeters satisfy

\begin{gathered} AE=2x+1,\\ EC=5x-11,\\ BE=6. \end{gathered}

Find $x$ and the full lengths $AC$ and $BD$ .

Lengths are in centimeters. Each diagonal is bisected separately.

Solution. The two parts of $AC$ have equal lengths.

\begin{aligned} 2x+1&=5x-11,\\ 12&=3x,\\ x&=4. \end{aligned}

Each part of $AC$ is $9$ cm. Each part of $BD$ is $6$ cm.

\begin{gathered} AC=18\text{ cm},\\ BD=12\text{ cm}. \end{gathered}

The unequal full lengths are consistent with both diagonals being bisected.

Check your understanding

In parallelogram $ABCD$ , diagonals meet at $E$ . Which pair of conclusions follows from diagonal bisection?

A The full diagonals must be congruent B All four segments from

E

must have equal lengths C The diagonals must meet at right angles D Each diagonal has its own two equal halves

Choose sides, angles, or diagonals. Predict which measurements will agree, then move D. The construction keeps opposite sides parallel; the written proofs explain the observations.

Move D, keep the opposite-side relationships

Compare sides

Scroll sideways to see the whole diagram.

Sideways shift Vertical separation Measurements

AB and CD: 6 units. BC and DA: about 4.47 units. Length measurements are rounded to two decimal places.

ABCD stays a parallelogram as D moves. Compare one family of measurements at a time; dragging illustrates the proved properties.

Turning works, while a fold can fail

A half-turn is a rotation through $180^\circ$ . It sends a point to the opposite side of its center, along the same line and at the same distance. See Rotations for the point-by-point definition.

Let $O$ be the intersection of the diagonals of a parallelogram. For a concrete instance, if $OA$ and $OC$ are both $8$ cm on opposite rays, a half-turn sends $A$ to $C$ ; it similarly exchanges the endpoints of a diagonal with $5$ cm halves. The bisection proof supplies exactly those equal distances for every parallelogram.

Primes name the rotated vertices. The two outlines coincide after a half-turn.

The half-turn symmetry of a parallelogram#

Diagonal bisection says that $O$ is the midpoint of $AC$ and of $BD$ . A half-turn about $O$ therefore exchanges $A$ with $C$ and $B$ with $D$ . A rotation sends a segment to the segment joining its endpoint images, so it exchanges each side with its opposite side. The entire boundary and its interior match the original parallelogram.

A half-turn does not imply a reflection. An oblique parallelogram has no right angles; a general oblique member with unequal adjacent sides has no reflection axes.

This proposed reflection fails: A′ is not a vertex of the original parallelogram.

Here is why no alternative axis rescues that general member. A reflection of a convex quadrilateral must preserve its four corners and their boundary adjacency. If it fixes a corner, its axis passes through that corner. It cannot fix both noncollinear incident sides individually, so it must swap them, forcing adjacent side lengths to be equal. That is excluded here.

If no corner is fixed and some corner is exchanged with a neighbor, those consecutive angles must be equal. Since they are supplementary, each would be $90^\circ$ , also excluded. The only remaining vertex exchange pairs both opposite vertices; its axis would have to be perpendicular to both crossing diagonals, which is impossible. These cases rule out every reflection for an oblique parallelogram with unequal adjacent sides.

Rectangles and rhombi can have additional reflection symmetries; a square has four reflection axes. Their full properties and proofs belong to the later lesson on special parallelograms. The word “general” matters: the no-reflection conclusion does not apply to all special members.

Worked example 5 Compare two proposed motions

Problem. A parallelogram has a $65^\circ$ corner and unequal adjacent sides of $6$ cm and $10$ cm. Its diagonals meet at $O$ .

Case A: Does the half-turn about $O$ map it onto itself? Explain, and count its rotational symmetries, including the identity.

Case B: Can reflection in a diagonal map it onto itself? Explain.

Solution. Case A: Yes. Diagonal bisection makes $O$ the midpoint of both diagonals, so the half-turn exchanges opposite vertices. Every symmetry preserves the diagonal intersection, so a nonidentity rotational symmetry has center $O$ . It must send a $65^\circ$ corner to the opposite $65^\circ$ corner: the neighboring corners are $115^\circ$ , and a nonidentity rotation fixes no corner. The opposite corners lie on opposite rays from $O$ , which forces a half-turn. With the identity included, there are $2$ rotational symmetries.

Case B: No. Reflection in either diagonal would fix its endpoints and exchange the two sides incident at each endpoint. Those incident lengths are $6$ cm and $10$ cm, so they cannot match.

Check your understanding

Which symmetry statement is true for every parallelogram, including special members?

A It has exactly zero reflection axes B It has a half-turn about its diagonal intersection C Either diagonal is a reflection axis D It has a quarter-turn about its diagonal intersection

Parallelograms

Learning goals

Start with the definition

A diagonal reveals opposite sides and angles

Each diagonal cuts the other in half

Turning works, while a fold can fail

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice

Your progress

Parallelograms

Learning goals

Start with the definition

A diagonal reveals opposite sides and angles

Consecutive angles share a straight-angle total

Each diagonal cuts the other in half

Turning works, while a fold can fail

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice

Your progress