Trapezoids and Kites

Watch the video explainer

Learning goals

Identify bases and justify isosceles tests using the inclusive definitions
Prove and apply trapezoid midsegment relationships
Compare reflection and rotational symmetries across trapezoid cases
Derive kite angle and diagonal properties from congruent triangles
Distinguish the guaranteed properties of a kite from the rhombus special case

Choose the bases before using a rule

A trapezoid is a quadrilateral with at least one pair of opposite parallel sides. Choose a parallel pair as its bases; the other two sides are its legs. A parallelogram is therefore a trapezoid, and either opposite pair may be chosen as its bases. In this lesson every named quadrilateral is convex, with vertices in boundary order.

For example, if the chosen bases of $ABCD$ are $AB$ and $CD$ , the legs are $AD$ and $BC$ . Calling them legs does not assert that they are nonparallel.

Relative to chosen parallel bases, an isosceles trapezoid has congruent angles at one base. A rectangle qualifies because both angles at any chosen base are right angles. In a parallelogram, consecutive angles are supplementary; they can be congruent only when both are right angles.

Worked example 1 Check the hypothesis on equal legs

Problem. An oblique parallelogram $ABCD$ has chosen bases $AB$ and $CD$ . Its angle at $A$ is $60^\circ$ , and its angle at $B$ is $120^\circ$ . Explain why its legs are congruent, and decide whether it is an isosceles trapezoid for these bases.

A counterexample must satisfy the proposed test while failing its conclusion.

Solution. Opposite sides of a parallelogram are congruent, so its two legs $AD$ and $BC$ are congruent. It is a trapezoid under our inclusive definition.

Its angles at base $AB$ are different. It is not an isosceles trapezoid. Equal legs alone are not a sufficient test when two opposite-side pairs may be parallel.

Check your understanding

Which extra hypothesis makes congruent legs a sufficient isosceles-trapezoid test?

A Exactly one opposite-side pair is parallel B Both opposite-side pairs are parallel C The diagonals intersect D The quadrilateral has four sides

Isosceles trapezoids come with a reflection

Equal angles at one base also force equal angles at the other. For instance, base angles of $68^\circ$ leave $112^\circ$ at each end of the other base: the angles along each leg are supplementary by parallel lines.

Why equal base angles give equal legs and diagonals#

Let $AB$ and $CD$ be the chosen bases, with congruent angles at $A$ and $B$ . Reflect in the perpendicular bisector of $AB$ . This swaps $A$ and $B$ . A line parallel to $AB$ maps onto itself under this reflection, so the line $CD$ stays in place.

The ray $AD$ maps to the ray from $B$ that makes the same base angle, on the same side of $AB$ . That ray is $BC$ . Their intersections with the line $CD$ must therefore correspond: $D$ maps to $C$ .

The reflection pairs the legs $AD$ and $BC$ , and also pairs the diagonals $AC$ and $BD$ . A reflection preserves lengths, so each pair is congruent. It also proves that this perpendicular bisector is a reflection axis of the whole trapezoid.

Reflection swaps A with B and D with C, so it also swaps the two diagonals.

The equal-legs test has a restricted converse. Suppose a convex trapezoid has exactly one opposite parallel pair and congruent legs. Its bases cannot have equal length, because one opposite pair both parallel and congruent would make it a parallelogram. Name the longer base $AB$ .

Why equal legs work with exactly one parallel pair#

Through $C$ , draw a parallel to $AD$ meeting $AB$ at $E$ . The quadrilateral $AECD$ is a parallelogram. Its opposite sides show that $AE$ has length $CD$ , so $E$ is strictly between $A$ and $B$ because $CD$ is shorter than $AB$ .

Its other opposite-side equality is

EC=AD.

The given congruent legs now make triangle $EBC$ isosceles, with equal sides $EC$ and $BC$ . Its base angles at $E$ and $B$ are congruent. Because $EC$ is parallel to $AD$ , its angle at $E$ equals the trapezoid’s angle at $A$ . Thus the two angles at base $AB$ are congruent.

The auxiliary parallel transfers a leg into triangle EBC. These equal-length marks belong to the proof.

Worked example 2 Use angle and diagonal information

Problem. A convex isosceles trapezoid $ABCD$ has chosen bases $AB$ and $CD$ . The angle at $A$ is $74^\circ$ . Its diagonals have lengths, in centimeters,

\begin{gathered} AC=4x+3\\ BD=6x-11. \end{gathered}

Find the angles at $B,C,D$ , the value of $x$ , and each diagonal length.

The text supplies the angle and lengths. The sketch shows the named diagonals.

Solution. The base angles at $A$ and $B$ are congruent. Each angle at the other base supplements one of them.

\begin{gathered} m\angle B=74^\circ\\ m\angle C=106^\circ\\ m\angle D=106^\circ. \end{gathered}

The diagonals are congruent, so

\begin{aligned} 4x+3&=6x-11\\ 14&=2x\\ x&=7. \end{aligned}

Substitution gives both requested lengths.

\begin{gathered} AC=31\text{ cm}\\ BD=31\text{ cm}. \end{gathered}

Join the leg midpoints

The segment joining the midpoints of a trapezoid’s legs is its midsegment. It is parallel to the bases, and its length is the average of their lengths. The average comes from two triangle midsegments, one contributing half of each base.

In a triangle, a segment joining two side midpoints is parallel to the third side and half as long. That result follows from similarity; revisit Triangle Proportionality and Midsegments if either conclusion needs a refresher.

Why the trapezoid midsegment averages the bases#

First take bases of $12$ cm and $26$ cm. Draw diagonal $AC$ and let $P$ be its midpoint. In triangle $ADC$ , the segment from leg midpoint $M$ to $P$ is parallel to $CD$ and has length $13$ cm. In triangle $ABC$ , the segment from $P$ to leg midpoint $N$ is parallel to $AB$ and has length $6$ cm. These are parts of the same straight segment, so the whole length is $19$ cm.

One diagonal turns the quadrilateral problem into two triangle-midsegment problems.

For any positive base lengths $a$ and $b$ , use the same construction. In a convex quadrilateral, $M$ and $N$ lie on opposite sides of diagonal $AC$ . The triangle theorem makes both $MP$ and $PN$ parallel to the bases. The unique parallel through $P$ puts them on one straight line, with $P$ between $M$ and $N$ .

\begin{gathered} MP=\frac b2\\ PN=\frac a2\\ MN=\frac{a+b}{2}. \end{gathered}

Nothing depended on the particular numbers. This proof also works for a parallelogram with chosen bases; it never required exactly one parallel pair.

Worked example 3 Recover a base

Problem. A convex trapezoid has bases of $9$ cm and $b$ cm. The segment joining its leg midpoints is $16$ cm long. Find $b$ , and state the segment’s direction relative to the bases.

A possible naming of the bases and leg midpoints.

Solution. A midsegment has the average base length.

\begin{aligned} 16&=\frac{9+b}{2}\\ 32&=9+b\\ b&=23. \end{aligned}

The other base is $23$ cm long. The midsegment is parallel to both bases. Its length lies between $9$ cm and $23$ cm, as an average should.

Check your understanding

A trapezoid has bases of 18 cm and 30 cm. What is its midsegment length?

A 12 cm B 24 cm C 48 cm D 15 cm

Compare the symmetry cases

A symmetry maps the whole figure onto itself. The identity, which leaves every point in place, counts as a rotational symmetry. Reflection axes are counted separately.

The dashed lines are reflection axes already justified by the relevant shape properties.

For an unequal-legged trapezoid with exactly one parallel pair, there is no reflection axis and the identity is its only rotational symmetry. A reflection would have to preserve the unique parallel pair. Because the bases have unequal lengths, it must fix each base and swap its endpoints, forcing the perpendicular bisector of both bases to be one axis. That would match the legs, contrary to their unequal lengths.

A matching rotation must also preserve the unique parallel pair. The unequal bases must each map to themselves. A nonidentity rotation fixing a segment is a half-turn about its midpoint, but the two bases have different midpoints. Thus only the identity can match the whole figure.

For a non-parallelogram isosceles trapezoid, that perpendicular axis exists by the reflection proof, and it is the only one. Its only rotational symmetry is the identity. A nonidentity rotation preserving a quadrilateral moves its vertices in a cycle; it can only be a half-turn or a quarter-turn. A half-turn would swap the unequal bases, and a quarter-turn would require two parallel pairs, so neither works.

For a nonsquare rectangle, there are two reflection axes through opposite side midpoints, and two rotational symmetries: the identity and a half-turn. A square has four reflection axes and four rotations, including the identity.

The inclusive definition also includes other parallelograms. A general oblique parallelogram with unequal adjacent sides has no reflection axis and has the identity and a half-turn. A nonsquare rhombus has two diagonal reflection axes and those same two rotations. These cases follow from Parallelograms and Rectangles, Rhombi, and Squares. An inclusive family need not have one symmetry count.

Check your understanding

Count rotational symmetries including the identity. Which statement is correct?

A Every trapezoid has two rotations B Every isosceles trapezoid has one reflection axis C A nonsquare rectangle has two reflection axes and two rotations D A square has two reflection axes

A kite has a distinguished diagonal

A kite is a convex quadrilateral with two disjoint pairs of adjacent congruent sides, using each side once. For example, sides of $5,5,8,8$ in boundary order can form a kite: pair the two $5$ s and pair the two $8$ s. Four equal sides qualify too, so rhombi and squares are kites.

Name the equal-side meeting vertices $A$ and $C$ :

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

The diagonal $AC$ joins those meeting vertices. It bisects the angles at $A$ and $C$ , and it is the perpendicular bisector of $BD$ . It also gives a reflection axis. The opposite angles at $B$ and $D$ are congruent.

The side-pair markings identify AC as the diagonal whose bisection properties must be proved.

The useful earlier fact is that SSS congruence matches corresponding triangle angles after three side pairs have been matched. The proof in Triangle Congruence explains why side evidence determines the matching.

Why the kite diagonal bisects and reflects#

Draw $AC$ . Triangles $ABC$ and $ADC$ have the two given matching side pairs and the common side $AC$ , so they are congruent by SSS. Their corresponding angles show that $AC$ bisects the vertex angles at $A$ and $C$ , and that the whole angles at $B$ and $D$ are congruent.

Let $O$ be the diagonal intersection. Convexity puts it inside both diagonals. Triangles $ABO$ and $ADO$ have equal sides $AB$ and $AD$ , the common side $AO$ , and congruent included angles because $AC$ bisects the angle at $A$ . SAS makes these triangles congruent.

It follows that

BO=DO.

Their angles at $O$ are congruent and form a linear pair, so each is a right angle. Thus $AC$ is the perpendicular bisector of $BD$ . Reflection in $AC$ swaps $B$ and $D$ and fixes $A$ and $C$ , so it maps the kite onto itself.

The bisection and right-angle marks are conclusions of the congruent-triangle proof.

Worked example 4 Read the correct halves

Problem. A convex kite $ABCD$ has equal adjacent pairs $AB,AD$ and $BC,CD$ . Its diagonals meet at $O$ . The lengths $AO,OC,BO$ are $9$ cm, $4$ cm, and $6$ cm, respectively. Find $DO$ and $BD$ . Decide whether $BD$ bisects $AC$ , and justify the decision.

The named lengths are givens; neither a midpoint nor a right angle is marked in the question figure.

Solution. The diagonal through the equal-side meeting vertices is $AC$ , so it bisects $BD$ .

\begin{gathered} DO=6\text{ cm}\\ BD=12\text{ cm}. \end{gathered}

The other diagonal $BD$ does not bisect $AC$ : the given parts of $AC$ are $9$ cm and $4$ cm, which are unequal. Perpendicular diagonals do not imply mutual bisection.

Check your understanding

In a convex kite ABCD, the equal pairs meet at A and C. Which conclusion is guaranteed?

A BD bisects AC B Both diagonals have equal length C AC bisects BD and the angles at A and C D The angles at A and C are congruent

The rhombus special case

An ordinary kite need not have a half-turn: in Example 4, a half-turn would swap $A$ and $C$ , but their distances from $BD$ are different. A kite that is not a rhombus has just the one diagonal reflection axis. Write its side lengths in boundary order as $p,p,q,q$ , with $p$ different from $q$ . A half-turn or quarter-turn would match a side of length $p$ to a side of length $q$ , so neither is a symmetry. The vertex between the two sides of length $p$ is unique, and so is the vertex between the two sides of length $q$ . Any reflection must fix these two vertices individually, so its axis must be their joining line $AC$ .

Worked example 5 Add the missing midpoint

Problem. A convex kite $ABCD$ has equal pairs meeting at $A$ and $C$ . Its diagonals meet at $O$ . In addition, $O$ is the midpoint of $AC$ . Prove that $ABCD$ is a rhombus.

Solution. The kite proof already gives that $O$ is the midpoint of $BD$ . Both diagonals now bisect each other, so the test from Proving a Quadrilateral Is a Parallelogram makes $ABCD$ a parallelogram. Its opposite sides are congruent, and its kite side pairs are adjacent and congruent. Together these make all four sides congruent, which is the rhombus condition.

In the special case, O also bisects AC. The parallelogram test then combines with the adjacent side pairs.

In a rhombus, either diagonal can be chosen through the equal-pair meeting vertices, so both bisection directions hold. A nonsquare rhombus has two diagonal reflection axes and a half-turn; a square has four axes and quarter-turns as well. Those extra symmetries belong to the special cases, not to every kite.

Check your understanding

Which extra statement makes a convex kite a rhombus, when AC joins its equal-side meeting vertices?

A The diagonals are perpendicular B AC bisects BD C The angles at B and D are congruent D BD also bisects AC

Trapezoids and Kites

Learning goals

Choose the bases before using a rule

Isosceles trapezoids come with a reflection

Join the leg midpoints

Compare the symmetry cases

A kite has a distinguished diagonal

The rhombus special case

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice

Your progress