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Complete Course of Mathematics
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Topic 16: Statistics & Probability


Content On This Page
Quadrilateral: Definition and Basic Terms Angle Sum Property of a Quadrilateral Types of Quadrilaterals (Trapezium, Kite, Parallelogram)
Special Types of Parallelograms (Rectangle, Rhombus, Square) Properties of a Parallelogram (Sides, Angles, Diagonals) Properties of a Rectangle, Rhombus, and Square
Properties of a Trapezium and Kite


Quadrilaterals: Introduction, Types, and Properties



Quadrilateral: Definition and Basic Terms

A quadrilateral is a closed figure formed by four line segments in a plane. It is a fundamental type of polygon with the following characteristics:

The word "quadrilateral" comes from the Latin words quadri (meaning 'four') and latus (meaning 'side').

A general quadrilateral ABCD showing vertices, sides, angles, and diagonals AC and BD.

Basic Terms Related to a Quadrilateral

Consider a general quadrilateral named ABCD, as shown in the image above. The key terms associated with it are:


Convex and Concave Quadrilaterals

Quadrilaterals can be broadly classified into two types based on their interior angles and the position of their diagonals:

Side-by-side comparison of a convex quadrilateral and a concave quadrilateral.

In most standard geometry discussions, when the term "quadrilateral" is used without further specification, it refers to a convex quadrilateral.



Angle Sum Property of a Quadrilateral

The Angle Sum Property of a Quadrilateral states that the sum of the measures of the four interior angles of any simple (non-self-intersecting) convex quadrilateral is always exactly $360^\circ$. This is a fundamental property derived from the Angle Sum Property of a Triangle.

Statement

For any convex quadrilateral ABCD, the sum of its interior angles $\angle A, \angle B, \angle C,$ and $\angle D$ is $360^\circ$. That is, $\angle A + \angle B + \angle C + \angle D = 360^\circ$.


Proof

Let's prove this property using the Angle Sum Property of a Triangle.

Given:

A convex quadrilateral ABCD.

To Prove:

$\angle A + \angle B + \angle C + \angle D = 360^\circ$.

Construction:

Draw a diagonal, say $\overline{AC}$, to divide the quadrilateral into two triangles.

Quadrilateral ABCD with diagonal AC drawn, dividing it into triangle ABC and triangle ADC.

Proof:

The diagonal $\overline{AC}$ divides the quadrilateral ABCD into two triangles: $\triangle ABC$ and $\triangle ADC$.

We know from the Angle Sum Property of a Triangle that the sum of the interior angles of any triangle is $180^\circ$.

Consider $\triangle ABC$:

$\angle BAC + \angle ABC + \angle BCA = 180^\circ$

... (i)

(Sum of angles in $\triangle ABC$)

Consider $\triangle ADC$:

$\angle DAC + \angle ADC + \angle ACD = 180^\circ$

... (ii)

(Sum of angles in $\triangle ADC$)

Now, add equation (i) and equation (ii):

$(\angle BAC + \angle ABC + \angle BCA) + (\angle DAC + \angle ADC + \angle ACD) = 180^\circ + 180^\circ$

Rearranging the terms, we group the angles that form the vertices of the quadrilateral:

$(\angle BAC + \angle DAC) + \angle ABC + (\angle BCA + \angle ACD) + \angle ADC = 360^\circ$

Observe the angles at vertices A and C in the figure:

  • The angle at vertex A of the quadrilateral is $\angle A$ or $\angle DAB$, which is the sum of the angles $\angle BAC$ and $\angle DAC$ from the two triangles: $\angle DAB = \angle BAC + \angle DAC$.
  • The angle at vertex C of the quadrilateral is $\angle C$ or $\angle BCD$, which is the sum of the angles $\angle BCA$ and $\angle ACD$ from the two triangles: $\angle BCD = \angle BCA + \angle ACD$.

Substitute these back into the rearranged sum of angles:

$\angle DAB + \angle ABC + \angle BCD + \angle ADC = 360^\circ$

Which can be written as:

$\angle A + \angle B + \angle C + \angle D = 360^\circ$

This proves that the sum of the interior angles of any convex quadrilateral is $360^\circ$.



Types of Quadrilaterals (Trapezium, Kite, Parallelogram)

Quadrilaterals are classified into different types based on the specific properties related to their sides, angles, and diagonals. Some of the common types of quadrilaterals include Trapeziums, Kites, and Parallelograms, which form the basis for further classifications.

1. Trapezium (or Trapezoid in US English)

Definition: A Trapezium is a quadrilateral in which at least one pair of opposite sides is parallel. In some definitions, a trapezium is required to have *exactly* one pair of parallel sides, but the broader definition (at least one pair) is also widely used, especially in higher mathematics and internationally.

A trapezium ABCD with AB parallel to DC.

In the figure above, quadrilateral ABCD is a trapezium because side $\overline{AB}$ is parallel to side $\overline{DC}$ ($AB \parallel DC$). The parallel sides are called the bases of the trapezium (e.g., $\overline{AB}$ and $\overline{DC}$). The non-parallel sides ($\overline{AD}$ and $\overline{BC}$ in this case) are called the legs or non-parallel sides.

Isosceles Trapezium

A special type of trapezium is the Isosceles Trapezium.

Definition: An Isosceles Trapezium is a trapezium where the non-parallel sides are equal in length.

An isosceles trapezium ABCD with AB parallel to DC and AD = BC.

In the figure, ABCD is an isosceles trapezium with $AB \parallel DC$ and $AD = BC$.

Properties of an Isosceles Trapezium:


2. Kite

Definition: A Kite is a quadrilateral that has two distinct pairs of equal-length adjacent sides.

A kite ABCD with adjacent sides AB=AD and CB=CD.

In the figure, quadrilateral ABCD is a kite because adjacent sides $\overline{AB}$ and $\overline{AD}$ are equal ($AB = AD$), and adjacent sides $\overline{CB}$ and $\overline{CD}$ are equal ($CB = CD$). Importantly, the two pairs of equal sides are *distinct*, meaning the length of a side from the first pair is generally not equal to the length of a side from the second pair (i.e., $AB \neq BC$).

Properties of a Kite:


3. Parallelogram

Definition: A Parallelogram is a quadrilateral in which both pairs of opposite sides are parallel.

A parallelogram ABCD with AB parallel to DC and AD parallel to BC.

In the figure, quadrilateral ABCD is a parallelogram because $\overline{AB} \parallel \overline{DC}$ and $\overline{AD} \parallel \overline{BC}$.

Parallelograms possess several key properties related to their sides, angles, and diagonals that distinguish them from other quadrilaterals. These properties are explored in detail in the next section.



Special Types of Parallelograms (Rectangle, Rhombus, Square)

Building upon the definition of a parallelogram (a quadrilateral with both pairs of opposite sides parallel), we encounter special cases that possess additional specific properties. These include the Rectangle, Rhombus, and Square. They are all parallelograms, but with extra conditions.

1. Rectangle

Definition: A Rectangle is a parallelogram with one angle being a right angle ($90^\circ$). Since a parallelogram has opposite angles equal and consecutive angles supplementary, if one angle is $90^\circ$, all four angles must be $90^\circ$. Thus, an alternative definition is a quadrilateral with four right angles.

A rectangle ABCD showing four right angles.

In rectangle ABCD, $\overline{AB} \parallel \overline{DC}$, $\overline{AD} \parallel \overline{BC}$, and $\angle A = \angle B = \angle C = \angle D = 90^\circ$.

Properties of a Rectangle:

A rectangle inherits all the properties of a parallelogram, plus:


2. Rhombus

Definition: A Rhombus is a parallelogram with all four sides equal in length. An alternative definition is a quadrilateral with all four sides equal in length.

A rhombus ABCD showing four equal sides AB=BC=CD=DA.

In rhombus ABCD, $\overline{AB} \parallel \overline{DC}$, $\overline{AD} \parallel \overline{BC}$, and $AB = BC = CD = DA$.

Properties of a Rhombus:

A rhombus inherits all the properties of a parallelogram, plus:


3. Square

Definition: A Square is the most special type of parallelogram. It can be defined in several equivalent ways:

A square ABCD showing four equal sides and four right angles.

In square ABCD, $\overline{AB} \parallel \overline{DC}$, $\overline{AD} \parallel \overline{BC}$, $AB = BC = CD = DA$, and $\angle A = \angle B = \angle C = \angle D = 90^\circ$.

Properties of a Square:

A square combines the properties of a parallelogram, a rectangle, and a rhombus. Its properties include:


Hierarchy and Relationships among Quadrilaterals

The relationship between these special quadrilaterals can be understood as a hierarchy:

This hierarchy can be summarised as:

Square $\subset$ Rectangle $\subset$ Parallelogram $\subset$ Quadrilateral

Square $\subset$ Rhombus $\subset$ Parallelogram $\subset$ Quadrilateral

Kite $\subset$ Quadrilateral (A kite is not necessarily a parallelogram)

Trapezium $\subset$ Quadrilateral (A trapezium is not necessarily a parallelogram, unless it has two pairs of parallel sides, in which case it is a parallelogram)



Properties of a Parallelogram (Sides, Angles, Diagonals)

A parallelogram is a special type of quadrilateral defined by the property that both pairs of its opposite sides are parallel. This fundamental definition gives rise to several key properties concerning its sides, angles, and diagonals.

Parallelogram ABCD with diagonals AC and BD intersecting at O.

Let ABCD be a parallelogram, which means $\overline{AB} \parallel \overline{DC}$ and $\overline{AD} \parallel \overline{BC}$.

Theorem 1: Opposite sides of a parallelogram are equal in length.

Given:

A parallelogram ABCD, with $\overline{AB} \parallel \overline{DC}$ and $\overline{AD} \parallel \overline{BC}$.

To Prove:

$AB = DC$ and $AD = BC$.

Construction:

Draw the diagonal $\overline{AC}$.

Proof:

Consider $\triangle ABC$ and $\triangle CDA$.

$\angle BAC = \angle DCA$

(Alternate interior angles, since $AB \parallel DC$ and AC is transversal) ... (1)

$\angle BCA = \angle DAC$

(Alternate interior angles, since $AD \parallel BC$ and AC is transversal) ... (2)

AC = CA

(Common side to both triangles) ... (3)

From (1), (2), and (3), by the ASA (Angle-Side-Angle) congruence criterion, we have:

$\triangle ABC \cong \triangle CDA$

Since the triangles are congruent, their corresponding parts are equal by CPCT (Corresponding Parts of Congruent Triangles).

AB = CD

BC = DA

Thus, the opposite sides of a parallelogram are equal in length.


Theorem 2: Opposite angles of a parallelogram are equal.

Given:

A parallelogram ABCD.

To Prove:

$\angle A = \angle C$ and $\angle B = \angle D$.

Proof:

We have already proved in Theorem 1 that $\triangle ABC \cong \triangle CDA$ by drawing the diagonal $\overline{AC}$.

By CPCT, the corresponding angles are equal:

$\angle ABC = \angle CDA$

This implies $\angle B = \angle D$.

Also, from the alternate interior angles used in the congruence proof:

$\angle BAC = \angle DCA$

$\angle DAC = \angle BCA$

Adding these two equalities:

$\angle BAC + \angle DAC = \angle DCA + \angle BCA$

From the figure, we know that $\angle BAC + \angle DAC = \angle DAB = \angle A$, and $\angle DCA + \angle BCA = \angle BCD = \angle C$.

Therefore,

$\angle A = \angle C$

Thus, the opposite angles of a parallelogram are equal.


Theorem 3: Consecutive angles (adjacent angles) of a parallelogram are supplementary.

(Note: Supplementary angles add up to $180^\circ$)

Given:

A parallelogram ABCD.

To Prove:

$\angle A + \angle B = 180^\circ$, $\angle B + \angle C = 180^\circ$, $\angle C + \angle D = 180^\circ$, and $\angle D + \angle A = 180^\circ$.

Proof:

In a parallelogram ABCD, we have $\overline{AD} \parallel \overline{BC}$. Consider the transversal $\overline{AB}$ intersecting these parallel lines.

The angles $\angle A$ and $\angle B$ are consecutive interior angles on the same side of the transversal. We know that consecutive interior angles formed by a transversal intersecting parallel lines are supplementary.

$\angle A + \angle B = 180^\circ$

(Consecutive interior angles, AD $\parallel$ BC and AB is transversal) ... (1)

Similarly, consider the other pairs of consecutive angles:

  • With $\overline{AB} \parallel \overline{DC}$ and transversal $\overline{BC}$: $\angle B + \angle C = 180^\circ$ (Consecutive interior angles)
  • With $\overline{AD} \parallel \overline{BC}$ and transversal $\overline{DC}$: $\angle D + \angle C = 180^\circ$ (Consecutive interior angles)
  • With $\overline{AB} \parallel \overline{DC}$ and transversal $\overline{AD}$: $\angle A + \angle D = 180^\circ$ (Consecutive interior angles)

Thus, all pairs of consecutive angles in a parallelogram are supplementary.


Theorem 4: The diagonals of a parallelogram bisect each other.

Definition of Bisect: To divide into two equal parts.

Given:

A parallelogram ABCD, with diagonals $\overline{AC}$ and $\overline{BD}$ intersecting at point O.

To Prove:

The diagonals bisect each other, meaning $AO = OC$ and $BO = OD$.

Proof:

Consider $\triangle AOB$ and $\triangle COD$.

Since ABCD is a parallelogram, $\overline{AB} \parallel \overline{DC}$.

Consider transversal $\overline{AC}$ intersecting $\overline{AB}$ and $\overline{DC}$. The alternate interior angles are equal:

$\angle OAB = \angle OCD$

(Alternate interior angles, AB $\parallel$ DC) ... (1)

Consider transversal $\overline{BD}$ intersecting $\overline{AB}$ and $\overline{DC}$. The alternate interior angles are equal:

$\angle OBA = \angle ODC$

(Alternate interior angles, AB $\parallel$ DC) ... (2)

From Theorem 1, the opposite sides of a parallelogram are equal:

AB = CD

(Opposite sides of parallelogram ABCD) ... (3)

From (1), (2), and (3), by the ASA (Angle-Side-Angle) congruence criterion (using $\angle OAB$, side AB, and $\angle OBA$ for $\triangle AOB$, and $\angle OCD$, side CD, and $\angle ODC$ for $\triangle COD$), we have:

$\triangle AOB \cong \triangle COD$

Since the triangles are congruent, their corresponding parts are equal by CPCT:

AO = CO

(CPCT) ... (4)

BO = DO

(CPCT) ... (5)

Therefore, the point O is the midpoint of both diagonals $\overline{AC}$ and $\overline{BD}$. This proves that the diagonals of a parallelogram bisect each other.


Converse Theorems of Parallelogram Properties

The converses of the properties discussed above are also true and are very useful in proving that a given quadrilateral is a parallelogram. A quadrilateral is a parallelogram if any one of the following conditions holds:



Properties of a Rectangle, Rhombus, and Square

Rectangles, rhombuses, and squares are specialized types of parallelograms. This means that they automatically possess all the fundamental properties of parallelograms (opposite sides are equal and parallel, opposite angles are equal, consecutive angles are supplementary, and diagonals bisect each other). In addition to these, they have unique properties that define them and distinguish them from a general parallelogram.

Properties of a Rectangle

A rectangle is a parallelogram with four right angles. Let's explore its properties.

Rectangle ABCD with diagonals AC and BD intersecting at O.

Inherited Parallelogram Properties:

Specific Properties of a Rectangle:

Proof that Diagonals of a Rectangle are Equal

Given:

A rectangle ABCD.

To Prove:

The diagonals $\overline{AC}$ and $\overline{BD}$ are equal in length, i.e., $AC = BD$.

Proof:

Consider $\triangle ABC$ and $\triangle DCB$.

AB = DC

(Opposite sides of a parallelogram are equal)

$\angle ABC = \angle DCB$

(Both are $90^\circ$, property of a rectangle)

BC = CB

(Common side)

Therefore, by the SAS (Side-Angle-Side) congruence criterion, we have:

$\triangle ABC \cong \triangle DCB$

By CPCT (Corresponding Parts of Congruent Triangles):

AC = DB

Thus, the diagonals of a rectangle are equal.

Since the diagonals of a rectangle are equal and they bisect each other (as it's a parallelogram), the four segments formed at the intersection are also equal ($AO = BO = CO = DO$).

Note that in a general rectangle, the diagonals do not necessarily bisect the angles at the vertices, nor are they necessarily perpendicular to each other, unless it is also a square.


Properties of a Rhombus

A rhombus is a parallelogram with all four sides equal in length. Let's explore its properties.

Rhombus ABCD with diagonals AC and BD intersecting perpendicularly at O.

Inherited Parallelogram Properties:

Specific Properties of a Rhombus:

Proof that Diagonals of a Rhombus are Perpendicular Bisectors of Each Other and Bisect Vertex Angles

We already know the diagonals bisect each other because a rhombus is a parallelogram. Let the diagonals $\overline{AC}$ and $\overline{BD}$ intersect at O.

Given:

A rhombus ABCD with diagonals $\overline{AC}$ and $\overline{BD}$ intersecting at O.

To Prove:

$\overline{AC} \perp \overline{BD}$ and the diagonals bisect the angles at the vertices (e.g., $\angle DAO = \angle BAO$).

Proof:

Consider $\triangle AOD$ and $\triangle COD$.

AD = CD

(All sides of a rhombus are equal)

AO = CO

(Diagonals of a parallelogram bisect each other, and a rhombus is a parallelogram)

OD = OD

(Common side)

Therefore, by the SSS (Side-Side-Side) congruence criterion:

$\triangle AOD \cong \triangle COD$

By CPCT:

$\angle AOD = \angle COD$

Since $\angle AOD$ and $\angle COD$ form a linear pair on the line $\overline{AC}$, their sum is $180^\circ$.

$\angle AOD + \angle COD = 180^\circ$

(Linear pair)

Substituting $\angle AOD = \angle COD$, we get:

$\angle AOD + \angle AOD = 180^\circ$

$2 \angle AOD = 180^\circ$

$\angle AOD = 90^\circ$

This proves that the diagonals intersect at right angles ($\overline{AC} \perp \overline{BD}$).

Also by CPCT from $\triangle AOD \cong \triangle COD$:

$\angle ADO = \angle CDO$

This shows that the diagonal $\overline{BD}$ bisects the angle at vertex D ($\angle ADC$). Similarly, considering other pairs of congruent triangles ($\triangle AOB \cong \triangle COB$, $\triangle AOB \cong \triangle AOD$, $\triangle BOC \cong \triangle DOC$), we can prove that the diagonals bisect the angles at all vertices.

Note that in a general rhombus, the diagonals are not necessarily equal in length unless it is also a square.


Properties of a Square

A square is a quadrilateral that is both a rectangle and a rhombus. Therefore, it inherits all the properties of a parallelogram, a rectangle, and a rhombus. This makes the square the most symmetric quadrilateral.

Square ABCD with diagonals AC and BD intersecting perpendicularly at O.

Properties of a Square:

In summary, the diagonals of a square are equal, are perpendicular bisectors of each other, and bisect the vertex angles into $45^\circ$ angles.


Summary Table of Diagonal Properties for Parallelograms and Special Types

The table below summarises the key diagonal properties for a general parallelogram and its special cases:

Property Parallelogram Rectangle Rhombus Square
Diagonals Bisect Each Other? Yes Yes Yes Yes
Diagonals Equal? No (Not necessarily) Yes No (Not necessarily) Yes
Diagonals Perpendicular? No (Not necessarily) No (Not necessarily) Yes Yes
Diagonals Bisect Vertex Angles? No (Not necessarily) No (Not necessarily) Yes Yes


Properties of a Trapezium and Kite

Beyond parallelograms, there are other significant types of quadrilaterals, such as trapeziums and kites, which possess unique properties derived from their specific definitions regarding side lengths and parallelism.

Properties of a Trapezium

Definition Reminder: A Trapezium is a quadrilateral with at least one pair of opposite sides parallel.

Trapezium ABCD with AB parallel to DC.

In a trapezium ABCD, let the parallel sides be $\overline{AB}$ and $\overline{DC}$ ($AB \parallel DC$). These parallel sides are called the bases of the trapezium. The non-parallel sides, $\overline{AD}$ and $\overline{BC}$ in this case, are called the legs or non-parallel sides.

General Properties of a Trapezium:


Properties of an Isosceles Trapezium

An Isosceles Trapezium is a trapezium where the non-parallel sides (legs) are equal in length.

If ABCD is an isosceles trapezium with $AB \parallel DC$, then $AD = BC$.

Isosceles Trapezium ABCD with AB parallel to DC, AD=BC.

In addition to the properties of a general trapezium, an isosceles trapezium has these specific properties:

Proof that Base Angles of an Isosceles Trapezium are Equal ($\angle D = \angle C$)

Given:

An isosceles trapezium ABCD, where $AB \parallel DC$ and $AD = BC$.

To Prove:

$\angle D = \angle C$.

Construction:

Draw a line through B parallel to AD, intersecting DC produced at E.

Proof:

Since $AB \parallel DC$ and $AD \parallel BE$ (by construction), quadrilateral ABED is a parallelogram.

In parallelogram ABED, opposite sides are equal:

AD = BE

But we are given that AD = BC.

Therefore, BE = BC. This means $\triangle BCE$ is an isosceles triangle.

In $\triangle BCE$, angles opposite to equal sides are equal:

$\angle BEC = \angle BCE$

(Angles opposite to equal sides BE and BC)

Since AD $\parallel$ BE and DC is a transversal, the corresponding angles are equal:

$\angle D = \angle BEC$

(Corresponding angles)

Combining the equalities, we get $\angle D = \angle BCE$.

Note that $\angle BCE$ is the same as $\angle BCD$, which is $\angle C$ of the trapezium.

Therefore, $\angle D = \angle C$. This proves that the base angles at the base DC are equal.

A similar construction or argument can be used to prove $\angle A = \angle B$.

Proof that Diagonals of an Isosceles Trapezium are Equal ($AC = BD$)

Given:

An isosceles trapezium ABCD, where $AB \parallel DC$ and $AD = BC$.

To Prove:

The diagonals $\overline{AC}$ and $\overline{BD}$ are equal in length, i.e., $AC = BD$.

Proof:

Consider $\triangle ADC$ and $\triangle BCD$.

AD = BC

(Given, legs are equal)

$\angle D = \angle C$

(Base angles of an isosceles trapezium are equal, proved above)

DC = CD

(Common side)

Therefore, by the SAS (Side-Angle-Side) congruence criterion, we have:

$\triangle ADC \cong \triangle BCD$

By CPCT (Corresponding Parts of Congruent Triangles):

AC = BD

Thus, the diagonals of an isosceles trapezium are equal in length.


Properties of a Kite

Definition Reminder: A Kite is a quadrilateral that has two distinct pairs of adjacent sides equal in length.

Kite ABCD with AB=AD, CB=CD, and diagonals AC and BD.

In a kite ABCD, suppose the pairs of equal adjacent sides are $\overline{AB} = \overline{AD}$ and $\overline{CB} = \overline{CD}$. The vertices between the equal sides (A and C in this case) are on one diagonal, and the vertices between the unequal sides (B and D) are on the other diagonal. The diagonal connecting the vertices between equal sides (AC) is the axis of symmetry of the kite.

Properties of a Kite:

Proof of Key Diagonal Properties of a Kite

Let ABCD be a kite with $AB = AD$ and $CB = CD$. Let the diagonals $\overline{AC}$ and $\overline{BD}$ intersect at O.

Given:

A kite ABCD with $AB = AD$ and $CB = CD$. Diagonals $\overline{AC}$ and $\overline{BD}$ intersect at O.

To Prove:

(i) $AC \perp BD$

(ii) BO = OD

(iii) $\angle BAC = \angle DAC$ and $\angle BCA = \angle DCA$

(iv) $\angle ABC = \angle ADC$

Proof:

(i) and (iii): Consider $\triangle ABC$ and $\triangle ADC$.

AB = AD

(Given)

CB = CD

(Given)

AC = AC

(Common side)

By the SSS (Side-Side-Side) congruence criterion:

$\triangle ABC \cong \triangle ADC$

By CPCT:

$\angle BAC = \angle DAC$

(i) ... (1)

$\angle BCA = \angle DCA$

(ii) ... (2)

Equations (1) and (2) prove property (iii): the diagonal $\overline{AC}$ bisects the angles $\angle A$ and $\angle C$.

Now consider $\triangle ABO$ and $\triangle ADO$.

AB = AD

(Given)

$\angle BAO = \angle DAO$

(From (1), AC bisects $\angle A$)

AO = AO

(Common side)

By the SAS (Side-Angle-Side) congruence criterion:

$\triangle ABO \cong \triangle ADO$

By CPCT:

BO = DO

(iii) ... (3)

$\angle AOB = \angle AOD$

(iv) ... (4)

Equation (3) proves property (ii): the diagonal $\overline{AC}$ bisects the diagonal $\overline{BD}$.

From equation (4), $\angle AOB$ and $\angle AOD$ are equal and form a linear pair on the line $\overline{BD}$.

$\angle AOB + \angle AOD = 180^\circ$

(Linear pair)

Substituting $\angle AOB = \angle AOD$ from (4):

$2 \angle AOB = 180^\circ$

$\angle AOB = 90^\circ$

This proves property (i): the diagonals are perpendicular to each other ($AC \perp BD$).

(iv): Consider $\triangle ABD$ and $\triangle CBD$.

AB = CB

(Wait, this is incorrect! It should be $AB=AD$ and $CB=CD$. Let's use $\triangle ABC$ and $\triangle ADC$ again.)

From the congruence $\triangle ABC \cong \triangle ADC$ proved above, by CPCT:

$\angle ABC = \angle ADC$

This proves property (iv): the pair of opposite angles between the unequal sides are equal.