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Rhombuses, kites and trapezia

Geometry and measurement : Module 21Years : 9-10

June 2011

PDF Version of module

Assumed knowledge

The material in this module is a continuation of the module, Parallelograms and Rectangles, which is assumed knowledge for the present module. Thus the present module assumes:

• Confidence in writing logical argument in geometry written as a sequence of steps, each justified by a reason.
• Ruler-and-compasses constructions.
• The four standard congruence tests and their application to:
• proving properties of and tests for isosceles and equilateral triangles,
• proving properties of and tests for parallelograms and rectangles.
• Informal experience with rhombuses, kites, squares and trapezia.

Motivation

Logical argument, precise definitions and clear proofs are essential if one is to understand mathematics. These analytic skills can be transferred to many areas in commerce, engineering, science and medicine but most of us first meet them in high school mathematics.

Apart from some number theory results such as the existence of an infinite number of primes and the Fundamental Theorem of Arithmetic, most of the theorems students meet are in geometry starting with Pythagoras’ theorem.

Many of the key methods of proof such as proof by contradiction and the difference between a theorem and its converse arise in elementary geometry.

As in the module, Parallelograms and Rectangles, this module first stresses precise definitions of each special quadrilateral, then develops some of its properties, and then reverses the process, examining whether these properties can be used as tests for that particular special quadrilateral. We have seen that a test for a special quadrilateral is usually the converse of a property. For example, a typical property−test pair from the previous module is the pair of converse statements:

• If a quadrilateral is a parallelogram, then its diagonals bisect each other.
• If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.

Congruence is again the basis of most arguments concerning rhombuses, squares, kites and trapezia, because the diagonals dissect each figure into triangles.

A number of the theorems proved in this module rely on one or more of the previous theorems in the module. This means that the reader must understand a whole ‘sequence of theorems’ to achieve some results. This is typical of more advanced mathematics.

In addition, two other matters are covered in these notes.

• The reflection and rotation symmetries of triangles and special quadrilaterals
  are identified and related to congruence.
• The tests for the kite also allow several important standard constructions to be explained very simply as constructions of a kite.

Content

Symmetries of triangles, parallelograms and rectangles

We begin by relating the reflection and rotation symmetries of isosceles triangles, parallelograms and rectangles to the results that we proved in the previous module, Paralleograms and Rectangles.

The axis of symmetry of an isosceles triangle

In the module, Congruence, congruence was used to prove that the base angles of an isosceles triangle are equal. To prove that B = C in the diagram opposite, we constructed the angle-bisector AM of the apex A, then used the SAS congruence test to prove that

ABM ACM

This congruence result, however, establishes much more than the equality of the base angles. It also establishes that the angle bisector AM is the perpendicular bisector of the base BC. Moreover, this fact means that AM is an axis of symmetry of the isosceles triangle.

These basic facts of about isosceles triangles will be used later in this module and in the module, Circle Geometry:

Theorem

In an isosceles triangle, the following four lines coincide:

• The angle bisector of the apex angle.
• The line joining the apex and the midpoint of the base.
• The line through the apex perpendicular to the base.
• The perpendicular bisector of the base.

This line is an axis of symmetry of the isosceles triangle. It has, as a consequence, the interesting property that the centroid, the incentre, the circumcentre and the orthocentre of ABC all tie on the line AM. In general, they are four different points. See the module, Construction for details of this.

Extension − Some further tests for a triangle to be isosceles

The theorem above suggests three possible tests for a triangle to be isosceles. The first two are easy to prove, but the third is rather difficult because simple congruence cannot be used in this ‘non-included angle’ situation.

The symmetries of an equilateral triangle

An equilateral triangle is an isosceles triangle in three different ways, so the three vertex angle bisectors form three axes of symmetry meeting each other at 60°. In an equilateral triangle, each vertex angle bisector is the perpendicular bisector of the opposite side − we proved in the previous module that in any triangle, these three perpendicular bisectors are concurrent. They meet at a point which is the centre of a circle through all three vertices. The point is called the circumcentre and the circle is called the circumcircle of the triangle.

An equilateral triangle is also congruent to itself in two other orientations:

ABC BCA CAB (SSS),

corresponding to the fact that it has rotation symmetry of order 3. The centre of this rotation symmetry is the circumcentre O described above, because the vertices are equidistant from it.

Other triangles do not have reflection or rotation symmetry

In a non-trivial rotation symmetry, one side of a triangle is mapped to a second side,
and the second side mapped to the third side, so the triangle must be equilateral.

In a reflection symmetry, two sides are swapped, so the triangle must be isosceles.

Thus a triangle that is not isosceles has neither reflection nor rotation symmetry. Such a triangle is called scalene.

Rotation symmetry of a parallelogram

Since the diagonals of a parallelogram bisect each other, a parallelogram has rotation symmetry of order 2 about the intersection of its diagonals. Joining the diagonal AC of a parallelogram ABCD produces two congruent triangles,

ABC CDA (AAS);

Reflection symmetry of a rectangle

A rectangle is a parallelogram, so it has rotationsymmetry of order 2 about the intersection of its diagonals. This is even clearer in a rectangle than in a general parallelogram because the diagonals have equal length, so their intersection is the circumcentre of the circumcircle passing through all four vertices.
The line through the midpoints of two opposite sides of a rectangle dissects the rectangle into two rectangles that are congruent to each other, and are in fact reflections of each other in the constructed line. There are two such lines in a rectangle, so a rectangle has two axes of symmetry meeting right angles.
It may seem obvious to the eye that the intersection of these two axes of symmetry is the circumcentre of the rectangle, which is intersection of the two diagonals. This is illustrated in the diagram to the right, but it needs to be proven.

 

Rhombuses

The Greeks took the word rhombos from the shape of a piece of wood that was whirled about the head like a bullroarer in religious ceremonies. This derivation does not imply a definition, unlike the words ‘parallelogram’ and ‘rectangle’, but we shall take their classical definition of the rhombus as our definition because it is the one most usually adopted by modern authors.

Definition of a rhombus

A rhombus is a quadrilateral with all sides equal.

First property of a rhombus − A rhombus is a parallelogram

Since its opposite sides are equal, a rhombus is a parallelogram − this was our second test for a parallelogram in the previous module. A rhombus thus has all the properties of a parallelogram:

• Its opposite sides are parallel.
• Its opposite angles are equal.
• Its diagonals bisect each other.
• It has rotation symmetry of order two about the intersection of its diagonals.

When drawing a rhombus, there are two helpful orientations that we can use, as illustrated below.

The rhombus on the left looks like a ‘pushed-over square’, and has the orientation we usually use for a parallelogram. The rhombus on the right has been rotated so that it looks like the diamond in a pack of cards. It is often useful to think of this as the standard shape of a rhombus.

Constructing a rhombus using the definition

It is very straightforward to construct a rhombus using the definition of a rhombus. Suppose that we want to construct a rhombus with side lengths 5cm and acute vertex angle 50°.

• Draw a circle with radius 5cm.
• Draw two radii OA and OB meeting at 50°
  at the centre O.
• Draw arcs with the same radius 5cm and centres
  A and B, and let P be their point of intersection.

The figure OAPB is a rhombus because all its sides are 5cm.

This leads to yet another way to construct a line parallel to a given line through a given point P.

• Choose any point A on the line .
• Draw an arc with centre A and radius AP
  cutting at B.
• Complete the rhombus PABQ as before.

Then PQ || because the figure PABQ is a rhombus.

Second property of a rhombus − Each diagonal bisects two vertex angles

Theorem

Each diagonal of a rhombus bisects the vertex angles through which it passes.

Proof

	Let ABCD be a rhombus with the diagonal AC drawn.
	Let	BAC	= α
		BAC	= α	(base angles of isosceles ABC)
	so	DAC	= α	(alternate angles, BC || AD):

That is AC bisects BAD

The axes of symmetry of a rhombus

The exercise above showed that each diagonal of a rhombus dissects the rhombus into two congruent triangles that are reflections of each other in the diagonal,

ABC ADC and BAD BCD.

Thus the diagonals of a rhombus are axes of symmetry.
The following property shows that these two axes are perpendicular.

Third property of a rhombus − The diagonals are perpendicular

The proof given here uses the theorem about the axis of symmetry of an isosceles triangle proven at the start of this module. Two other proofs are outlined as exercises.

Theorem

The diagonals of a rhombus are perpendicular.

Proof

Let ABCD be a rhombus,
with diagonals meeting at M.

To prove that AC ⊥ BD.

By the previous theorem, AM is the angle bisector of DAB.

Hence AM ⊥ BD, because A is the apex of the isosceles triangle ABD,

The diagonals also bisect each other because a rhombus is a parallelogram, so we usually state the property as

‘The diagonals of a rhombus bisect each other at right angles.’

We now turn to tests for a quadrilateral to be a rhombus. This is a matter of establishing that a property, or a combination of properties, gives us enough information for us to conclude that such a quadrilateral is a rhombus.

First test for a rhombus − A parallelogram with two adjacent sides equal

We have proved that the opposite sides of a parallelogram are equal, so if two adjacent sides are equal, then all four sides are equal and it is a rhombus.

Theorem

If two adjacent sides of a parallelogram are equal, then it is a rhombus.

This test is often taken as the definition of a rhombus.

Second test for a rhombus − A quadrilateral whose diagonals bisect each other
at right angles

Theorem

A quadrilateral whose diagonals bisect each other at right angles is a rhombus.

Proof

Let ABCD be a quadrilateral whose diagonals bisect
each other at right angles at M.

We prove that DA = AB. It follows similarly that

AB = BC and BC = CD

AMB AMD (SAS)

So AB = AD and by the first test above ABCD is a rhombus.

A quadrilateral whose diagonals bisect each other is a parallelogram, so this test is often stated as

‘If the diagonals of a parallelogram are perpendicular, then it is a rhombus.’

This test gives us another construction of a rhombus.

• Construct two perpendicular lines intersecting
  at M.
• Draw two circles with centre M and different radii.
• Join the points where alternate circles cut the lines.

This figure is a rhombus because its diagonals bisect each other at right angles.

 

 

Third test for a rhombus − A quadrilateral in which the diagonals bisect
the vertex angles

Theorem

If each diagonal of a quadrilateral bisects the vertex angles through which it passes, then the quadrilateral is a rhombus.

Proof

Let ABCD be a quadrilateral, and suppose the diagonals bisect the angles, then let

DAC = BAC = α ABD = CBD = β

BCA = DCA = γ CDB = ADB = δ

To prove that ABCD is a rhombus.

First,		2α + 2β + 2γ + 2δ = 360°		(angle sum of quadrilateral ABCD)
		α + β + γ + δ = 180°
Secondly,		α + 2β + γ = 180°		(angle sum of ABC):
Combining these, β = δ,
Hence		AB || DC and BC || AD		(alternate angles are equal)
and		ABCD is a parallelogram
and		AD = AB		(opposite angles are equal in ABD);

so ABCD is a rhombus because it is a parallelogram with a pair of adjacent sides equal.

Extension − Quadrilaterals whose diagonals are perpendicular

The converse of a property is not necessarily a test. For example, a quadrilateral with perpendicular diagonals need not be a rhombus − just place two sticks across each other at right angles and join their endpoints. The following exercise gives an interesting characterisation of quadrilaterals with perpendicular diagonals.

Both parts of the proof are applications of Pythagoras’ theorem. One half is straightforward, the other requires proof by contradiction and an ingenious construction.

Squares

We usually think of a square as a quadrilateral with all sides equal and all angles right angles. Now that we have dealt with the rectangle and the rhombus, we can define a square concisely as:

Definition of a square

A square is a quadrilateral that is both a rectangle and a rhombus.

Properties of a square

A square thus has all the properties of a rectangle, and all the properties of a rhombus.

• Opposite sides are parallel.
• The diagonals meet each side at 45°.
• The diagonals are equal in length, and bisect each other at right angles.
• The two diagonals, and the two lines joining the midpoints of opposite sides, are axes of symmetry.

Symmetries of a square

The intersection of the two diagonals is the circumcentre of the circumcircle through all four vertices. We have already seen, in the discussion of the symmetries of a rectangle, that all four axes of symmetry meet at the circumcentre.

A square ABCD is congruent to itself in three other orientations,

ABCD BCDA CDAB DABC

corresponding to the fact that it has rotation symmetry of order 4. The centre of the rotation symmetry is the circumcentre, because the vertices are equidistant from it.

Constructing a square

The most obvious way to construct a square of side length 6cm is to construct a right angle, cut off lengths of 6cm on both arms with a single arc, and then complete the parallelogram.

 

Alternatively, we can combine the previous diagonal constructions of the rectangle of the rhombus. Construct two perpendicular lines intersecting at O, draw a circle with centre O, and join up the four points where the circle cuts the lines.

 

 

Extension − A dissection problem

The following problem requires the construction that divides a given interval in a given ratio − see the module Constructions.

Kites

Some of the distinctive properties of the diagonals of a rhombus hold also in a kite, which is a more general figure. Because of this, several important constructions are better understood in terms of kites than in terms of rhombuses.

Definition of a kite

A kite is a quadrilateral with two pairs of adjacent equal sides.

A kite may be convex or non-convex, as shown in the diagrams above. Only the convex cases are presented in the proofs below − the non-convex cases are similar, but are left
as exercises.

Constructing a kite using its definition

The definition allows a straightforward construction using compasses. Suppose that we want to construct a kite with side lengths 8cm and 5cm, with the two 8cm sides meeting at 60°.

• Draw a circle with centre O and radius 8cm.
• Draw two radii OA and OB meeting at O at an angle of 60°.
• Complete the kite OAPB by drawing circles of radius 5cm with centres at A and B.

The last two circles meet at two points P and P0, one inside the large circle and one outside, giving a convex kite and a non-convex kite meeting the specifications.

Notice that the reflex angle of a non-convex kite is formed between the two shorter sides.

First property of a kite − The axis of symmetry

Theorem

Let ABCD be a kite with AB = AD and CB = CD.

aABC ADC.

bThe diagonal AC is an axis of reflection symmetry of the kite.

c The axis AC bisects the vertex angles at A and C.

d B = D.

Proof

The congruence follows from the definition, and the other parts follow from the congruence.

Second property of a kite − The axis is the perpendicular bisector of the other diagonal

Theorem

The axis of a kite is the perpendicular bisector of the other diagonal.

Proof

Let ABCD be a kite with AB = AD and CB = CD.

and let the diagonals of the kite meet at M.

Using the theorem about the axis of symmetry of an isosceles triangle, the bisector AM of the apex angle of the isosceles triangle ABD is also the perpendicular bisector of its base BD.

Hence BM = MD and AM ⊥ BD

So AC ⊥ BD

 

Tests for a kite

The converses of some these properties of a kite are tests for a quadrilateral to be a kite.

Theorem

If one diagonal of a quadrilateral bisects the two vertex angles through which it passes, then the quadrilateral is a kite.

 

 

 

Theorem

If one diagonal of a quadrilateral is the perpendicular bisector of the other diagonal, then the quadrilateral is a kite.

 

 

 

Kites and geometric constructions

Three of the most common ruler-and-compasses constructions can be explained in terms of kites.

1. The first diagram to the right shows the construction of the angle bisector of XOY. This construction works because the axis OP of the kite OAPB bisects the vertex angle at O.

Notice that the radii of the arcs meeting at P need not be the same as the radius of the first arc with centre O.

2. The second diagram to the right shows the construction of the perpendicular to a line from a point P. This construction works because the diagonals of the kite PAQB are perpendicular.

Notice that the radii of the arcs meeting at Q need not be the same as the radii of the original arc with centre P.

3. The two diagrams below show the construction of the perpendicular bisector of AB. This construction works because the axis PQ of the kite APBQ bisects the other diagonal AB at right angles.

In the diagram to the left, the radii of the arcs meeting at P are not the same as the radii of the arcs meeting at Q. Of course it is usual in this construction, and far more convenient, to use equal radii − as in the diagram to the right − in which case the figure constructed is a rhombus.

Trapezia

Trapezia also have a characteristic property involving the diagonals, but the property concerns areas, not lengths or angles.

Definition of a trapezium

A trapezium is a quadrilateral with one pair of opposite sides parallel.

Although the name is Latin (the plural is ‘trapezia’), it originally comes from the Greek word trapeza, meaning ‘table’. The figure is called a ‘trapezoid’ in the USA.

The angles of a trapezium

Using co-interior angles, we can see that a trapezium has two pairs of adjacent supplementary angles.

 

Conversely, if a quadrilateral is known to have one pair of adjacent supplementary angles, then it is a trapezium.

The diagonals of a trapezium

The diagonals of a convex quadrilateral dissect the quadrilateral into four triangular regions, as shown in the diagrams below. In a trapezium, two of these triangles have the same area, and the converse of this property is a test for a quadrilateral to be a trapezium.

These results are written as exercises because they are not usually regarded as standard theorems for students to know.

Extension − Isosceles trapezia

The trapezia that occur in this exercise are called isosceles trapezia. Further results about isosceles trapezia can be found at https://en.wikipedia.org/wiki/Isosceles_trapezoid.

Links forward

This module completes the study of special quadrilaterals using congruence. Similarity is a generalisation of congruence, and when it has been developed, some further results about special quadrilaterals will become possible. The module, Circle Geometry will use some special quadrilaterals, and will also introduce cyclic quadrilaterals, which are quadrilaterals whose four vertices all lie on the one circle − they will be the last special quadrilaterals discussed in these modules.

All triangles have both a circumcircle and an incircle. The only quadrilaterals that have a circumcircle are those with opposite angles supplementary, the situation with incircles is interesting. For example, a rhombus always has an incircle.

As an easy exercise show that if the lengths of the diagonals of the rhombus are p and q and the radius of the incircle is r then

A kite has an incircle as well but its radius is difficult to calculate.

A kite is determined by the triangle with side lengths a, b and included angle θ. If the lengths of the diagonals are p and q show that :

2pq = ab sin θ

When complex numbers are graphed on Argand diagrams, many arithmetic and algebraic results are proved or illustrated using special quadrilaterals. In particular if z1 and z2 are two complex numbers of equal modulus then the four numbers, z1, z1+ z2 and z2 form a rhombus so, as a consequence, z1 + z2 and z1,− z2 are perpendicular vectors.

History and applications

This illustrates very well the constant attitude in mathematics that an investigation is not complete until a theorem with a true converse has been identified. It reminds us too that logic, accompanied by the intuition of diagrams, should always be a strong motivation in geometry.

Whenever a surface is divided up, triangles and special quadrilaterals are involved, particularly when parallel lines are used in the dissection. Thus surveyors analysing suburban blocks or farming lots will try to use the simplest geometric shapes in their analysis, and architects, who often have great freedom to invent striking patterns for their building, often use special quadrilaterals other than simple squares and rectangles in their designs. Infinite tilings of the plane, for example, are possible with any other quadrilateral. Trigonometry is also an essential part of these processes, and trigonometry and geometry should be seen as a unit rather than as two disconnected topics. Several exercises in these modules have required such connections to be made.

A standard problem for computer programmers is to encrypt pictures with as little storage as possible, and they typically divide up the picture into simple geometrical shapes as part of that process − a recent study uses interlocking trapezia for this purpose.

All the special quadrilaterals of this and the previous module, apart from the kite, were studied by the ancient Greeks as part of their systematic investigation of geometry. The kite was named and brought to attention in modern times, partly because that it clarifies several important geometric constructions, but also because it demonstrates that some of the properties of a rhombus hold in more general quadrilaterals.

ANSWERS TO EXERCISES

EXERCISE 1

a This is a simple application of the SAS congruence test.

 

 

b This is a simple application of the AAS congruence test.

 

 

c Suppose AC < AB. Choose P on the internal AB so that AP = AC and join PM

			PAM CAM (SAS)
	so		APM = C = γ		(matching angles of congruent triangles)
	so		BPM = 180° − γ		(straight angle)
	Also		PM = CM		(matching sides of congruent triangles)
	So		MPB = MBP = 180° − γ		(base angles of isosceles MPB)
	Hence		α + α + γ + (180° − γ ) = 180°		(angle sum of ABC)
			α = 0°. So AC = AB

EXERCISE 2

In the triangles APM and CQM in the given diagram:

1	AP = QC		(each is half the opposite side of a rectangle)
2	MAP = MCQ		(alternate angles, AB || DC)
3	AMP = CMQ		(vertically opposite angles)
	so	APM CQM		(AAS).
	Hence	AM = CM	(matching sides of congruent triangles).
	That is PQ bisects AC

EXERCISE 3

AB2	= 52 + 52 − 2 × 5 × 5 × cos 50°		PO2	= 52 + 52 + 2 × 5 × 5 × cos 50°
	= 50 (1 − cos 50°)			= 50 (1 + cos 50°)
AB	≈ 4.23cm,		PO	≈ 9.06cm,

EXERCISE 4

The congruence ABC ADC follows by the SSS test.

EXERCISE 5

a ABM ADM (SAS or SSS).

b Let α = BAM and β = ABM.

	Then		CBM = β		(previous property)
	and		BCM = α		(base angles of isosceles ABC);
	so		2α + 2β = 180°		(angle sum of ABC)
			α + β = 90°
	Hence		AMB = 90°		(angle sum of ABM).

or A rhombus is a parallelogram. So BM = MD. Hence AMB ADM (SSS)
Thus AMB = AMD and the diagonals are perpendicular.

EXERCISE 6

Using Pythagoras’ theorem, the side length is 2cm. Using trigonometry, the vertex angles are about 67.38° and 112.62°.

EXERCISE 7

First, let ABCD be a convex quadrilateral whose diagonals meet at right angles at M.
Let the sides and the intervals on the diagonals have lengths as on the diagram to the right. Then using Pythagoras’ theorem,

a2 + c2	= (p2 + q2) + (r2 + s2)
	= (q2 + r2) + (p2 + s2)
	= b2 + d2, as required.

Conversely, let ABCD be a convex quadrilateral in which the diagonals are not perpendicular. The diagram will be as drawn on the right or its reflection, and it will be sufficient to consider only the one case. Let the lengths be as given on the figure, where
x ≠ 0 because AC is not perpendicular to BD. Using Pythagoras’ theorem,

			a2 + c2 = p2 + q2 + r2 + s2
	and		b2 + d2 = (q + x) 2 + r2 + (s + x)2 + p2
	so		(b2 + d2) − (a2 + c2) = 2x(q + s + x), which is not zero because x ≠ 0.
	Hence if a2 + b2 = c2 + d2 then AC ⊥ BD as required

EXERCISE 8

Using Pythagoras’ theorem, the required radius is 3cm.

EXERCISE 9

a Construct points P on BC and Q on DC so that

BP : PC = DQ : QC = 2 : 1

Then the triangles APB and APC have the same altitude AB,

and their bases are in the ratio 2 : 1, so

area APB : area APC = 2 : 1

Similarly area AQD : area AQC = 2 : 1. Since the diagonal AC divides the square into two congruent regions of equal area, the lines AP and AQ dissect the square into three regions of equal area.

or show that the area of each region is

b A similar argument works

EXERCISE 10

Take the convex case first. Using equilateral triangles and Pythagoras’ theorem, the axis is

4 + 3cm and the other diagonal is 8cm. Using trigonometry, the angle at the other end of the axis is about 106.26°, and the other two angles are each about 60° + 36.87°.

In the non-convex case, the axis is 4 − 3cm and the other diagonal is still 8cm. The angle at the other end of the axis is about 253.74°, and the other two angles are each about 60° − 36.87°.

EXERCISE 11

a Since the axis AC bisects the vertex angle at A,

ABM ADM (SAS);

from which it follows that AMB = AMD = 90°.

b	Let		BAM = DAM = α		(the axis bisects the vertex angle),
	and let		ABM = ADM = β		(base angles of isosceles ABD),
	Then		2α + 2β = 180°		(angle sum of ABD)
			α + β = 90°
	Hence		AMB = 90°		(angle sum of AMB).

EXERCISE 12

Using the AAS congruence test, ABC ADC.

So AB = AD, BC = DC and the quadrilateral is a kite.

EXERCISE 13

Using the SAS congruence test, ABM ADM and CBM CDM.

So AB = AD and CB = CD and the quadrilateral is a kite.

EXERCISE 14

Diagram: Kite 10

If one of the equal angles is included by the given sides, then there is no reason for the figure to be a kite, as is illustrated in the diagram on the left above.

If neither equal angle is included by the equal sides, then the figure is a kite. Be careful, however, because joining the diagonal AC in the diagram on the right above would not give a congruence proof because the angle in each triangle would be non-included.

Instead, join the other diagonal BD, as in the diagram on the right above.

Then		ABD = ADB	(base angles of isosceles ABD),
so		CBD = CDB	(subtracting equal angles from equal angles)
so		BC = DC	(opposite angles are equal in BCD)
		ABCD is a kite

EXERCISE 15

In the diagram to the right, join BD and let DBC = θ.

Then		CDB = θ		(base angles of isosceles CBD)
so		ABD = θ		(alternate angles, AB || DC)
so		ADB = θ		(base angles of isosceles ABD)
Hence		AD || BC		(alternate angles are equal)

so ABCD is a rhombus, because it is a parallelogram with two adjacent sides equal.

EXERCISE 16

The triangles ABC and DBC have the same perpendicular height and the same base BC, so area ABC = area DBC.

Subtracting the triangle MBC from both regions,

area AMB = area DMC.

EXERCISE 17

Adding MBC to both regions,

area ABC = area DBC

These two triangles have the same base BC, so they have the same perpendicular height. Hence AD || BC.

EXERCISE 18

a		Let		B=β		and construct P on BC so that DP || AB.
		Then		DP = AB		(opposite sides of a parallelogram)
		and		DPC = β		(corresponding angles, AB || DP).
		Suppose first that AB = DC.
		Then		PD = DC
		so		C = β		(base angles of isosceles DPC)
		Conversely, suppose that C = β
		Then		DC = DP		(opposite angles of DPC are equal)
		so		DC = AB

b If FG is an axis of symmetry, then reflection
in FG maps AB to DC, so AB = DC.

Conversely, suppose that AB = DC.

Then B = C by part a.

Drop perpendiculars AL and DM to the line BC.

		Then		AL = DM		(opposite sides of rectangle ALMD)
		so		ABL DMC		(AAS)
		Hence		BL = CM		(matching sides of congruent triangles)
		so		LG = MG		(subtract equal lengths from equal lengths)
		so		LG = AF		(opposite sides of a rectangle)
		Hence		FG ⊥ BC

so reflection in FG swaps B and C, and swaps A and D, as required.

c Since AD ≠ BC, a reflection cannot swap AD and BC.

If a reflection swapped AD and AB, then it would also swap BC and DC, so ABCD would be a kite with parallel sides, so it would be a parallelogram. Similarly a reflection cannot swap AD and DC.

 

The Improving Mathematics Education in Schools (TIMES) Project 2009-2011 was funded by the Australian Government Department of Education, Employment and Workplace Relations.

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