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Proving a Quadrilateral Is a Parallelogram: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 The side record

Convex quadrilateral $ABCD$ has side lengths in centimeters.
$\begin{gathered}AB=15\\CD=15\\BC=9\\AD=9\end{gathered}$
Decide whether these measurements guarantee a parallelogram. Name the sufficient test, or explain why none is complete.

The markings represent the two pairs in the measurement record.

Text description of this figure

Convex quadrilateral ABCD has vertices named in boundary order. The first and third sides have matching single ticks. The second and fourth sides have matching double ticks.
1. Hint 1
  Match each side with the side opposite it, using boundary order.
2. Hint 2
  A complete side test requires an equality in each opposite-side pair.
Answer

Yes; both opposite-side pairs are congruent.

Full solution

The first opposite-side pair has $AB=CD$ .

The second opposite-side pair has $BC=AD$ .

Both opposite-side pairs are congruent, so the side converse proves $ABCD$ is a parallelogram.
This conclusion uses exact givens rather than how the drawing looks.

Answer

Yes; both opposite-side pairs are congruent.

Key idea

Both opposite-side equalities complete a sufficient test.
Problem 2 Two crossing records

In each separate case, $ABCD$ is convex and its diagonals $AC,BD$ meet at $E$ . Decide whether the evidence proves a parallelogram, excludes a parallelogram, or leaves both possibilities. Give a reason. All lengths use the same unit.

Case A:
$\begin{gathered}AE=3\\EC=3\\BE=8\\ED=8\end{gathered}$
Case B:
$\begin{gathered}AE=4\\EC=4\\BE=6\\ED=9\end{gathered}$
1. Hint 1
  Check each diagonal separately at the common crossing.
2. Hint 2
  A complete midpoint test proves a parallelogram; failure of a necessary midpoint condition excludes one.
Answer

Case A: proves a parallelogram. Case B: excludes a parallelogram.

Full solution

Case A: the two parts of $AC$ match, and the two parts of $BD$ match.
Both diagonals bisect each other, so the quadrilateral is a parallelogram.

Case B: $BD$ is not bisected because its parts have lengths $6$ and $9$ .
Every parallelogram has both diagonals bisected, so this quadrilateral is not a parallelogram.

Answer

Case A: proves a parallelogram. Case B: excludes a parallelogram.

Key idea

Incomplete evidence leaves a question open, but a known violation of a necessary property settles it negatively.
Problem 3 A paper construction

Draw two nonparallel lines through a point $E$ . On the first line, place $A,C$ on opposite sides of $E$ , each $5$ cm from $E$ . On the second, place $B,D$ on opposite sides of $E$ , each $7$ cm from $E$ . Connect the four endpoints in boundary order $A,B,C,D$ .

Mark only the given equalities. Explain why the constructed quadrilateral is convex and why it must be a parallelogram. Does the argument require the two lines to be perpendicular?
1. Hint 1
  Think about how four rays from one crossing alternate around the crossing.
2. Hint 2
  Check the midpoint of each diagonal, and keep unequal whole diagonals distinct from unequal halves.
Answer

The alternating endpoints form a convex parallelogram; perpendicular lines are not required.

Full solution

The two nonparallel lines make four rays in alternating order.
Taking one point on each ray and joining them in that order gives a simple convex quadrilateral: each vertex lies between the neighboring ray directions, and each interior angle is less than a straight angle.

The construction gives the midpoint equalities
$\begin{gathered}AE=EC\\BE=ED\end{gathered}$

Thus both diagonals bisect each other, proving a parallelogram.
No step required a right angle between the diagonals.

The unequal diagonals share a midpoint; no right angle is required.

Answer

The alternating endpoints form a convex parallelogram; perpendicular lines are not required.

Key idea

A shared midpoint forces a parallelogram for any two nonparallel diagonal directions.
Problem 4 A proof from two side pairs

Convex $ABCD$ satisfies
$\begin{gathered}AB=CD\\AD=BC\end{gathered}$
Prove that both opposite-side pairs are parallel. Use diagonal $BD$ and triangle congruence, rather than citing the parallelogram side test as a finished theorem.

Only the given side equalities are marked.

Text description of this figure

Convex quadrilateral ABCD has vertices named in boundary order. The first and third sides have matching single ticks. The second and fourth sides have matching double ticks.
1. Hint 1
  Compare the triangles formed by the specified diagonal.
2. Hint 2
  Write the vertex correspondence before selecting the corresponding angle pairs.
3. Hint 3
  Use one alternate interior-angle pair to prove each parallel pair.
Answer

$\begin{gathered}AB\parallel CD\\AD\parallel BC\end{gathered}$

Full solution

The given side matches and the common side $BD$ give
$\triangle ABD\cong\triangle CDB$
by SSS.

Corresponding angles give
$\angle ABD\cong\angle CDB.$
They are alternate interior angles for $AB$ and $CD$ with transversal $BD$ , so $AB\parallel CD$ .

A second corresponding pair gives
$\angle ADB\cong\angle CBD.$
They are alternate interior angles for $AD$ and $BC$ , again with transversal $BD$ , so $AD\parallel BC$ .

Diagonal BD supplies the shared side for the proof.

Answer

$\begin{gathered}AB\parallel CD\\AD\parallel BC\end{gathered}$

Key idea

Triangle correspondence converts side equalities into the angle equalities needed for parallelism.
Problem 5 One angle pair

Construct a convex quadrilateral $ABCD$ with congruent opposite angles $A,C$ that is not a parallelogram. Give all four interior-angle measures and justify both the existence of your example and the failed parallelogram conclusion.
1. Hint 1
  A counterexample must meet the stated equality but can make the other opposite pair unequal.
2. Hint 2
  Try joining two congruent right triangles on opposite sides of a shared hypotenuse.
3. Hint 3
  A taught special right triangle can provide exact angles for the construction.
Answer

One example has angles $A,C$ of $90^\circ$ , angle $B$ of $60^\circ$ , and angle $D$ of $120^\circ$ , made by joining two congruent $30^\circ$ - $60^\circ$ - $90^\circ$ triangles along their hypotenuse.

Full solution

Take two congruent $30^\circ$ - $60^\circ$ - $90^\circ$ triangles with common hypotenuse $BD$ , on opposite sides of $BD$ .
Place their right-angle vertices at $A,C$ so each triangle has its $30^\circ$ angle at $B$ and its $60^\circ$ angle at $D$ .

The outer boundary is convex, with right angles at $A,C$ .
Adding the two triangle angles at each hypotenuse endpoint gives
$\begin{gathered}m\angle B=60^\circ\\m\angle D=120^\circ\end{gathered}$

The example satisfies the required opposite-angle equality at $A,C$ .
It is not a parallelogram because its other opposite angles are unequal, contrary to a necessary parallelogram property.

Answer

One example has angles $A,C$ of $90^\circ$ , angle $B$ of $60^\circ$ , and angle $D$ of $120^\circ$ , made by joining two congruent $30^\circ$ - $60^\circ$ - $90^\circ$ triangles along their hypotenuse.

Key idea

One opposite-angle equality is weaker than the complete two-pair test.
Problem 6 From angle matches to directions

In convex $ABCD$ , both opposite-angle pairs are congruent. Write $p$ for the common measure of angles $A,C$ , and $q$ for the common measure of angles $B,D$ . Prove both opposite-side pairs parallel without citing the opposite-angle parallelogram test.

Opposite angles carry matching arc patterns.

Text description of this figure

Convex quadrilateral ABCD has vertices named in boundary order. Opposite angles have matching arc counts, one arc at the first and third vertices and two at the second and fourth.
1. Hint 1
  The quadrilateral angle sum relates the two common measures.
2. Hint 2
  Use the resulting consecutive-angle sum along two different sides as transversals.
Answer

$\begin{gathered}p+q=180^\circ\\AD\parallel BC\\AB\parallel CD\end{gathered}$

Full solution

The four angles have measures $p,q,p,q$ in boundary order.
The quadrilateral angle sum gives
$\begin{aligned}2p+2q&=360^\circ\\p+q&=180^\circ\end{aligned}$

Angles $DAB$ and $ABC$ are same-side interior angles for lines $AD,BC$ along transversal $AB$ .
Their sum is $180^\circ$ , so the same-side interior-angle converse proves $AD\parallel BC$ .

Angles $ABC$ and $BCD$ are same-side interior angles for lines $AB,CD$ along transversal $BC$ .
Their sum is also $180^\circ$ , so $AB\parallel CD$ .

Answer

$\begin{gathered}p+q=180^\circ\\AD\parallel BC\\AB\parallel CD\end{gathered}$

Key idea

A supplementary consecutive-angle sum must be attached to the correct lines and transversal.
Problem 7 The first unsupported line

In convex $ABCD$ , diagonals $AC,BD$ meet at $E$ . The only given length equality is $AE=EC$ . Examine this attempted proof. Identify the first line that does not follow, explain the gap, and state one extra length equality that would repair the proof. Then explain how the repaired proof establishes both parallel pairs.

Line 1: $\angle AEB\cong\angle CED$ because they are vertical angles.

Line 2: $\triangle AEB\cong\triangle CED$ by SAS.

Line 3: $AB=CD$ because corresponding parts of congruent triangles are congruent.

Line 4: $AB\parallel CD$ from corresponding alternate interior angles.
1. Hint 1
  Check the inputs to the first triangle-congruence claim.
2. Hint 2
  SAS needs two side pairs around the included angle, while only one half-diagonal equality was supplied.
3. Hint 3
  After repairing that input, compare a second pair of triangles for the other side pair.
Answer

Line 2 is the first unsupported line. Add $BE=ED$ ; the two SAS comparisons then prove both opposite-side pairs parallel.

Full solution

Line 1 is valid for any two intersecting lines.
Line 2 is unsupported: SAS has the vertical-angle pair and the given side match
$AE=EC.$

SAS still needs the other side match
$BE=ED.$

With that extra equality, SAS gives
$\triangle AEB\cong\triangle CED.$
Corresponding angles $BAE,DCE$ are congruent alternate interior angles along $AC$ , proving
$AB\parallel CD.$

The other vertical-angle pair and the same two half-diagonal equalities give
$\triangle BEC\cong\triangle DEA$
by SAS.
Corresponding angles $BCE,DAE$ are congruent alternate interior angles along $AC$ , proving
$BC\parallel AD.$

The repaired proof has two midpoint equalities, shown by separate tick patterns.

Answer

Line 2 is the first unsupported line. Add $BE=ED$ ; the two SAS comparisons then prove both opposite-side pairs parallel.

Key idea

A correct vertical-angle statement does not supply a missing side equality for SAS.
Problem 8 A stronger incomplete record

Construct a convex quadrilateral whose whole diagonals are both $10$ cm long and whose crossing bisects $AC$ but does not bisect $BD$ . Give the four diagonal-part lengths, explain how to draw the quadrilateral, and prove it is not a parallelogram.
1. Hint 1
  Equal whole lengths still allow different divisions at the crossing.
2. Hint 2
  Choose equal halves on one line and unequal positive parts with the same total on the other.
3. Hint 3
  Place the lines perpendicular if that makes the construction easier; join endpoints in alternating order.
Answer

One construction on two perpendicular lines through $E$ has
$\begin{gathered}AE=5\text{ cm}\\EC=5\text{ cm}\\BE=7\text{ cm}\\ED=3\text{ cm}\end{gathered}$
It is not a parallelogram.

Full solution

Draw perpendicular lines meeting at $E$ .
Put $A,C$ on opposite rays of one line, each $5$ cm from $E$ .
Put $B,D$ on opposite rays of the other, respectively $7$ cm and $3$ cm from $E$ .
Joining them in alternating order forms a convex quadrilateral.

The whole diagonals have the same length:
$\begin{gathered}AC=5+5\\AC=10\\BD=7+3\\BD=10\end{gathered}$

The crossing is the midpoint of $AC$ but not of $BD$ , whose two parts have unequal lengths.
Since every parallelogram has both diagonals bisected, this quadrilateral is not a parallelogram.

Answer

One construction on two perpendicular lines through $E$ has
$\begin{gathered}AE=5\text{ cm}\\EC=5\text{ cm}\\BE=7\text{ cm}\\ED=3\text{ cm}\end{gathered}$
It is not a parallelogram.

Key idea

Congruent whole diagonals cannot replace a missing midpoint condition.
Problem 9 One pair carries two facts

Convex $ABCD$ satisfies
$\begin{gathered}AB\parallel CD\\AB=CD\end{gathered}$
Prove $AD\parallel BC$ using diagonal $BD$ . Name the triangle congruence and the angle pair that gives the parallelism; do not simply quote the one-pair parallelogram test.

The same opposite-side pair is marked parallel and congruent.

Text description of this figure

Convex quadrilateral ABCD has vertices named in boundary order. The first and third sides have matching single ticks. That same opposite-side pair has matching parallel arrows.
1. Hint 1
  The given parallel sides provide an angle pair along the chosen diagonal.
2. Hint 2
  The equal sides and the shared diagonal enclose that angle pair.
3. Hint 3
  Use the correspondence to obtain alternate interior angles on the other side pair.
Answer

$\begin{gathered}\triangle ABD\cong\triangle CDB\\\angle ADB\cong\angle CBD\\AD\parallel BC\end{gathered}$

Full solution

The given parallel lines $AB,CD$ , cut by $BD$ , give congruent alternate interior angles $ABD,CDB$ .
The side equality $AB=CD$ and shared diagonal surround those angles.

Thus
$\triangle ABD\cong\triangle CDB$
by SAS.
Corresponding angles give
$\angle ADB\cong\angle CBD.$

These are alternate interior angles for $AD,BC$ along transversal $BD$ .
Their converse proves $AD\parallel BC$ , completing the definition together with the given parallel pair.

SAS across BD supplies the missing alternate interior-angle pair.

Answer

$\begin{gathered}\triangle ABD\cong\triangle CDB\\\angle ADB\cong\angle CBD\\AD\parallel BC\end{gathered}$

Key idea

The one-pair test works because the parallel and congruent facts feed the same SAS comparison.
Problem 10 Keep the smallest record

Convex $ABCD$ is known to satisfy $AB\parallel CD$ . You may request any of the following three evidence cards.

Card P: $AB=CD$ .

Card Q: $AD=BC$ .

Card R: $AC=BD$ .

Find the minimum number of cards needed to guarantee a parallelogram, and list every single card that suffices. Prove your choice is minimal. Also decide whether cards Q and R together suffice, justifying your decision with exact geometric evidence.
1. Hint 1
  Keep track of which pair is already known parallel.
2. Hint 2
  For a rejected record, look for a quadrilateral with unequal parallel bases that still meets every requested equality.
3. Hint 3
  A centered upper base lets two right-triangle calculations verify both the sloping sides and the diagonals.
Answer

The minimum is one card, and P is the only sufficient single card. Q and R together are insufficient.

Full solution

Card P puts congruence on the same pair already known parallel.
The one-pair test therefore proves a parallelogram with one card.

To test all the weaker choices, construct parallel horizontal bases of lengths $14$ cm and $4$ cm, centered one above the other, with perpendicular separation $12$ cm.
Each horizontal offset is $5$ .
Joining corresponding endpoints gives a convex quadrilateral.

Each sloping side is a right-triangle hypotenuse with legs $5,12$ .
$\begin{gathered}5^2+12^2=169\\AD=13\\BC=13\end{gathered}$
Thus card Q holds.

Each diagonal has horizontal run $9$ and vertical rise $12$ .
$\begin{gathered}9^2+12^2=225\\AC=15\\BD=15\end{gathered}$
Thus card R holds too.

The opposite bases have different lengths, so this is not a parallelogram.
It shows that no cards, Q alone, R alone, and even Q with R leave a non-parallelogram possible.
Hence one card is minimal, and P is the only sufficient single card.

Answer

The minimum is one card, and P is the only sufficient single card. Q and R together are insufficient.

Key idea

Minimal sufficient evidence must force the conclusion, and every weaker proposed record needs a valid counterexample.

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