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Lesson 10 of 13

Quadrilaterals · Lesson 10 of 13

Kites

“Define kites and use their perpendicular-bisector and angle-bisector properties.”

Learning Objectives

• Recognise two pairs of equal adjacent sides. • Identify the special diagonal of a kite. • Prove its angle-bisection and perpendicular-bisection properties. • Construct a kite from diagonal lengths.

When two matching scalene triangles are reflected across a shared edge, their outline can resemble a flying kite. Look closely at its sides: two equal sides meet at one corner, and another equal pair meets at the opposite corner. This arrangement differs from the equal opposite sides of a parallelogram. Keeping track of where the equal sides meet will tell you which diagonal has the useful symmetry properties.

Label the kite carefully before applying a rule. One diagonal bisects the other, but a general kite does not require the bisection to work in both directions.

4.6 Kite and Trapezium

Kite

Name the vertices ABCD so that AB = BC and CD = DA. The first equal pair meets at B; the second meets at D. The diagonal BD joins those two meeting points.

The two pair lengths may be different. For example, AB = BC = 6 cm and CD = DA = 9 cm satisfies the condition without having four equal sides.

Definition
Kite

A quadrilateral that can be labelled ABCD with AB = BC and CD = DA: two non-overlapping pairs of equal adjacent sides.

OABCDAB = BC; AD = CD; BD perpendicularly bisects AC
Kite ABCD: BD is the special diagonal

Draw BD. Triangles ABD and CBD are congruent by SSS: AB = CB, AD = CD, and BD is common. Therefore ∠ABD = ∠CBD and ∠ADB = ∠CDB. BD bisects the angles at B and D.

Let O be the intersection of AC and BD in this convex kite. Triangles ABO and CBO have AB = CB, shared BO and equal included angles at B. They are congruent by SAS. Thus AO = CO and ∠AOB = ∠COB.

A, O and C lie on a straight line, so those equal angles total 180° and must each be 90°. This proves that BD is perpendicular to AC and passes through its midpoint. It does not prove BO = DO.

Special diagonal of the labelled kiteLaTeX
Using the bisected diagonal

Problem
AC = 18 cm in a kite with AB = BC and AD = CD. Find AO and OC.

A bisected vertex angle

Problem
The angle at B is 86°. Find ∠ABD and ∠DBC.

Figure it Out

Constructing a kite with 6 cm and 8 cm diagonals

Problem
Make a convex kite whose diagonals AC and BD have lengths 6 cm and 8 cm.

3 cm3 cm5 cm3 cmDACBOAC is bisected; BD need not be. Schematic.
One-way bisection in a kite
Names can overlap

A rhombus has four equal sides, so it also satisfies this kite definition. A square is a special rhombus and therefore also a kite under the same definition. A later comparison lesson will flag a conflicting answer-key statement explicitly.

Quiz

Quick check

A kite has:

Quick check

In kite ABCD with AB = BC and AD = CD, what is always true of BD?

Quick check

The special kite diagonal meets the other diagonal at:

Quick check

A rhombus is a special kite.

Quick check

If BD bisects an 86° kite angle, each part is:

Practice Problems

Practice Problems
  1. AB = BC = 9 cm and CD = DA = 6 cm. Classify ABCD and name its special diagonal.
  2. AC = 18 cm and BD perpendicularly bisects it at O. Find AO and OC.
  3. In that labelled kite, ∠ABC = 120°. Find ∠ABD and ∠DBC.
  4. Construct a convex kite with diagonals 6 cm and 8 cm that is not a rhombus.
  5. Why is every rhombus a kite, but not every kite a rhombus?
Practice 1: worked solution

1. There are two equal adjacent-side pairs, meeting at B and D. 2. The quadrilateral is a kite, and BD is its special diagonal.

Practice 2: worked solution

1. Bisection divides AC into equal halves. 2. AO = OC = 18 ÷ 2 = 9 cm.

Practice 3: worked solution

1. BD bisects the vertex angle at B. 2. Each part is 120° ÷ 2 = 60°.

Practice 4: worked solution

1. Draw AC = 6 cm and its perpendicular bisector through midpoint O. 2. Place B and D on opposite sides of AC with BO = 3 cm and DO = 5 cm; join the endpoints. 3. BD = 8 cm and bisects AC, so the side pairs are equal. Since AC does not bisect BD, the figure is not a rhombus.

Practice 5: worked solution

1. Four equal sides provide both equal adjacent-side pairs, so a rhombus meets the kite definition. 2. A kite may instead have pair lengths 6 cm and 9 cm. 3. It then has the required adjacent pairs but not four equal sides, so it is not a rhombus.

Key Takeaways

Key Takeaways

• Kites have two pairs of equal adjacent sides. • For AB = BC and AD = CD, BD is the special diagonal. • BD bisects the angles at B and D and perpendicularly bisects AC. • A general kite does not require AC to bisect BD. • Rhombuses, including squares, satisfy the inclusive kite definition. • To construct a convex kite, place the special-diagonal endpoints on opposite sides of the other diagonal.