Proving a Quadrilateral Is a Parallelogram

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Learning goals

Distinguish necessary properties from sufficient evidence that a quadrilateral is a parallelogram
Prove four parallelogram tests using triangle congruence and parallel-line converses
Choose a complete test and solve for values that satisfy its hypotheses
Construct counterexamples showing why incomplete side or diagonal evidence is insufficient

Evidence has a direction

A parallelogram is a quadrilateral with both pairs of opposite sides parallel. The preceding lesson proved that its opposite sides and angles are congruent and its diagonals bisect each other. Those statements start with a known parallelogram; this lesson starts with less information and proves the shape is one.

A necessary condition is something every parallelogram must have. A sufficient condition is enough to guarantee that a quadrilateral is a parallelogram. A condition can be both, but reversing a proven statement requires another proof.

For example, having one congruent opposite-side pair is necessary. A parallelogram must have it, yet a quadrilateral with only that evidence might have unequal sides in the other pair. We need a complete test.

Throughout this lesson, $ABCD$ is a convex quadrilateral with vertices in boundary order. Its sides do not cross, its vertices are distinct, and no three consecutive vertices lie on one line. We will use equal alternate interior angles, or supplementary same-side interior angles, to prove lines parallel. These are the converses established in Proving Lines Parallel or Perpendicular: identify the transversal and the angle positions before using them.

Check your understanding

Every parallelogram has congruent opposite sides. What must be established before using the reverse statement as a test?

A A picture that looks like a parallelogram B A proof that the given equalities force both opposite-side pairs parallel C The perimeter of one example D A measurement of one angle

Both opposite-side pairs are congruent

Two pairs of equal opposite sides force the parallel directions. Imagine side lengths $8,5,8,5$ around a convex quadrilateral. Draw one diagonal: the two triangles have the same three side lengths. Their matching angles will prove the needed parallelism.

Equal opposite sides and one shared diagonal provide an SSS comparison.

Both opposite-side pairs congruent imply a parallelogram#

Suppose the first opposite-side pair has equal lengths, and so does the second:

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

Draw $AC$ . The triangles $ABC$ and $CDA$ have the two given side matches and the shared side $AC$ . By SSS,

\triangle ABC\cong\triangle CDA.

Corresponding angles give the first pair of equal alternate interior angles:

\angle BAC\cong\angle DCA.

With transversal $AC$ , this proves $AB\parallel CD$ . The other matching pair is

\angle BCA\cong\angle DAC.

Again using $AC$ as transversal proves $BC\parallel AD$ . Both parallel pairs required by the definition have now been proved.

The side lengths $8$ and $5$ were only an illustration. The SSS argument used the equalities, not their particular values. Thus both opposite-side pairs congruent is a sufficient test.

Worked example 1 Make both side pairs match

Problem. Convex $ABCD$ has the following side lengths in centimeters.

\begin{gathered} AB=2x+1,\\ CD=11,\\ BC=x+2,\\ AD=7. \end{gathered}

Find every value of $x$ for which the opposite-side congruence test applies, and decide whether it proves a parallelogram.

Solution. The first pair requires

\begin{aligned} 2x+1&=11,\\ 2x&=10,\\ x&=5. \end{aligned}

The second pair requires

\begin{aligned} x+2&=7,\\ x&=5. \end{aligned}

Both conditions hold at $x=5$ . The side lengths are positive, and both opposite-side pairs are congruent, so $ABCD$ is a parallelogram. Solving only the first equation would leave half the test unchecked.

The completed side record has two congruent opposite-side pairs.

Both opposite-angle pairs are congruent

Two equal opposite-angle pairs also force parallel sides. For instance, if one pair measures $68^\circ$ each, the other pair must total $224^\circ$ , so each of those two equal angles is $112^\circ$ . Every consecutive pair then totals $180^\circ$ .

The arc patterns identify two separate opposite-angle pairs.

Both opposite-angle pairs congruent imply a parallelogram#

First try equal opposite angles of $74^\circ$ . The other pair has total

360^\circ-2(74^\circ)=212^\circ.

Because its two angles are equal, each is $106^\circ$ . A $74^\circ$ angle and a consecutive $106^\circ$ angle are supplementary. The same-side interior-angle converse then gives each parallel pair.

For the general argument, write $p$ for the common measure at $A,C$ and $q$ for the common measure at $B,D$ . The quadrilateral angle sum gives

\begin{aligned} 2p+2q&=360^\circ,\\ p+q&=180^\circ. \end{aligned}

Along transversal $AB$ , angles $DAB$ and $ABC$ are same-side interior angles for lines $AD$ and $BC$ . Their supplementary measures prove $AD\parallel BC$ .

Along transversal $BC$ , angles $ABC$ and $BCD$ are same-side interior angles for lines $AB$ and $CD$ . Their supplementary measures prove $AB\parallel CD$ . The particular angle measures did not matter; equality in each opposite pair made every consecutive sum $180^\circ$ .

Worked example 2 Use angle evidence

Problem. A convex quadrilateral has angles $A$ and $C$ of $83^\circ$ each. Angles $B$ and $D$ are congruent. Find the other two measures and prove that the quadrilateral is a parallelogram.

Solution. Remove the known pair from the quadrilateral total:

360^\circ-2(83^\circ)=194^\circ.

The remaining angles are equal, so

\begin{gathered} m\angle B=97^\circ,\\ m\angle D=97^\circ. \end{gathered}

Both opposite-angle pairs are congruent. The test just proved guarantees that the quadrilateral is a parallelogram.

Each arc pattern marks one of the completed opposite-angle pairs.

Check your understanding

Convex ABCD has congruent angles A and C. Which extra fact completes the opposite-angle test?

A Angles A and B are congruent B One diagonal is long C Angles B and D are congruent D One side is horizontal

The diagonals bisect each other

If the crossing is the midpoint of each diagonal, the quadrilateral is forced to be a parallelogram. For example, diagonal parts $4,4$ on one line and $7,7$ on the other give two side matches around each vertical-angle pair. SAS supplies the matching triangles.

Each equality compares parts of the same diagonal. No equality between the two diagonals is required.

Diagonals that bisect each other imply a parallelogram#

Let $E$ be the intersection of $AC$ and $BD$ . Suppose

\begin{gathered} AE=EC,\\ BE=ED. \end{gathered}

Angles $AEB$ and $CED$ are vertical angles, so they are congruent. The two given side equalities surround those angles. SAS gives

\triangle AEB\cong\triangle CED.

Corresponding angles $BAE$ and $DCE$ are congruent. They are alternate interior angles along $AC$ , proving $AB\parallel CD$ .

For the other side pair, angles $BEC$ and $DEA$ are vertical angles. The same half-diagonal equalities give

\triangle BEC\cong\triangle DEA

by SAS. Corresponding angles $BCE$ and $DAE$ are alternate interior angles along $AC$ , proving $BC\parallel AD$ .

The word bisect means divide into two equal parts. Both equalities are needed: saying that $AC$ bisects $BD$ gives only the equality on $BD$ .

Worked example 3 Check the common crossing

Problem. In convex $ABCD$ , diagonals $AC$ and $BD$ meet at $E$ . Their parts have the following lengths in centimeters.

\begin{gathered} AE=3t-2,\\ EC=t+6,\\ BE=9,\\ ED=2t+1. \end{gathered}

Find every value of $t$ that makes both diagonals bisect each other. Explain the classification.

Solution. On diagonal $AC$ ,

\begin{aligned} 3t-2&=t+6,\\ 2t&=8,\\ t&=4. \end{aligned}

On diagonal $BD$ ,

\begin{aligned} 9&=2t+1,\\ 8&=2t,\\ t&=4. \end{aligned}

At $t=4$ , the halves of $AC$ are each $10$ cm and the halves of $BD$ are each $9$ cm. Both diagonal equalities hold at the same intersection, so $ABCD$ is a parallelogram. The two whole diagonals need not have equal lengths.

The completed record has a midpoint on each of the unequal diagonals.

Check your understanding

In the midpoint proof, why are angles AEB and CED congruent before any parallel sides have been proved?

A The opposite sides are already parallel B All four half-diagonal lengths must be equal C They are vertical angles at the diagonal crossing D The drawing looks centered

The rods are pinned at their midpoints. Predict what kind of quadrilateral their endpoints form when the rods turn, then compare the drawing with the proof above.

Turn the rods without moving their shared midpoint

Join the endpoints in order A, B, C, D

Scroll sideways to see the whole diagram.

Angle between rods Common turn

AO and OC: 3 units. BO and OD: 2 units. Angle AOB: 70°. Common turn: 15° counterclockwise. The rods share midpoint O; neither rod lies on the other.

Rods AC and BD share midpoint O. Their endpoints are joined in boundary order A, B, C, D; the shared midpoint is the hypothesis of the diagonal test.

One opposite-side pair is parallel and congruent

A single opposite-side pair can be enough when that same pair has both properties. If $AB$ and $CD$ are parallel and each is $9$ units long, a diagonal gives an included angle from the parallel lines and two side matches for SAS.

Both the parallel arrows and the equal-length ticks belong to the same opposite-side pair.

One opposite-side pair parallel and congruent implies a parallelogram#

Suppose one opposite-side pair has both properties:

\begin{gathered} AB\parallel CD,\\ AB=CD. \end{gathered}

Draw diagonal $AC$ . Its alternate interior angles satisfy

\angle BAC\cong\angle DCA.

The given side equality and shared side $AC$ surround these equal angles. SAS gives

\triangle BAC\cong\triangle DCA.

Thus corresponding angles $BCA$ and $DAC$ are congruent. They are alternate interior angles for $BC$ and $AD$ , with transversal $AC$ , so $BC\parallel AD$ . Together with the given parallel pair, this proves a parallelogram.

Worked example 4 Keep the evidence on one pair

Problem. Convex $PQRS$ has one known parallel pair and these side lengths in inches.

\begin{gathered} PQ\parallel RS,\\ PQ=4u-3,\\ RS=17. \end{gathered}

Find a value of $u$ that guarantees a parallelogram, and justify it without assuming the other pair parallel.

Solution. Make the already parallel sides congruent:

\begin{aligned} 4u-3&=17,\\ 4u&=20,\\ u&=5. \end{aligned}

At $u=5$ , $PQ$ and $RS$ are each $17$ inches. That same opposite-side pair is parallel and congruent, so the proved test supplies the other parallel pair.

The same pair carries both types of evidence.

What incomplete evidence leaves open

To refute an incomplete test, construct one quadrilateral that satisfies all its givens and fails its conclusion. A sketch alone is not verification: use exact lengths, a construction, or an established theorem.

The parallel pair and the congruent pair are different pairs.

Here $AB$ and $CD$ are parallel, but their lengths are $12$ and $6$ . Each sloping side is the hypotenuse of a right triangle with legs $3$ and $4$ , so each has length $5$ . This quadrilateral has one parallel pair and a different congruent pair. It is not a parallelogram, because parallelograms have congruent opposite sides and the bases here are unequal.

The crossing bisects AC, but it does not bisect BD.

Here $AE$ and $EC$ are each $6$ , but $BE$ is $5$ and $ED$ is $3$ . Such a convex quadrilateral exists: draw two perpendicular lines through $E$ and put the four vertices at those distances. It is not a parallelogram, since every parallelogram has both diagonals bisected. One midpoint is not enough.

Worked example 5 Choose the missing fact

Problem. Treat these as separate cases about convex $ABCD$ . Decide which additional fact completes the indicated test in each case, and explain why.

Case A: $AB\parallel CD$ is given. Choose between the following two possible additions to complete the one-pair test.

AB=CD

AD=BC.

Case B: $AE=EC$ is given, where diagonals $AC$ and $BD$ meet at $E$ . Choose between the following two possible additions to complete the bisection test.

BE=ED

AC=BD.

Solution. Case A: choosing congruence of the already parallel pair completes the one-pair test. Congruence of the other opposite pair alone can hold in the unequal-base counterexample above.

Case B: $BE=ED$ completes diagonal bisection. Congruent whole diagonals do not say that $E$ is the midpoint of $BD$ ; they do not complete this test.

Check your understanding

A convex quadrilateral has one pair of opposite sides parallel, and its other pair congruent. What follows from only these facts?

A It must be a parallelogram B It cannot be a parallelogram C Its diagonals must be congruent D It may be a parallelogram, but the evidence does not guarantee one

Proving a Quadrilateral Is a Parallelogram

Learning goals

Evidence has a direction

Both opposite-side pairs are congruent

Both opposite-angle pairs are congruent

The diagonals bisect each other

One opposite-side pair is parallel and congruent

What incomplete evidence leaves open

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice

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