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:
- It has exactly four sides.
- It has exactly four vertices (corner points).
- It has exactly four interior angles.
- It is a two-dimensional shape.
The word "quadrilateral" comes from the Latin words quadri (meaning 'four') and latus (meaning 'side').

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:
- Vertices: These are the points where the adjacent sides of the quadrilateral meet. In quadrilateral ABCD, the vertices are A, B, C, and D.
- Sides: These are the four line segments that form the boundary of the quadrilateral. In quadrilateral ABCD, the sides are $\overline{AB}$, $\overline{BC}$, $\overline{CD}$, and $\overline{DA}$.
- Angles: These are the four interior angles formed at the vertices by the meeting of adjacent sides. In quadrilateral ABCD, the interior angles are $\angle A$ (or $\angle DAB$), $\angle B$ (or $\angle ABC$), $\angle C$ (or $\angle BCD$), and $\angle D$ (or $\angle CDA$).
- Diagonals: These are the line segments that connect opposite vertices of the quadrilateral. In quadrilateral ABCD, the diagonals are $\overline{AC}$ and $\overline{BD}$. A quadrilateral always has two diagonals.
- Adjacent Sides: Two sides of a quadrilateral are called adjacent if they share a common vertex. In quadrilateral ABCD, the pairs of adjacent sides are: ($\overline{AB}$, $\overline{BC}$), ($\overline{BC}$, $\overline{CD}$), ($\overline{CD}$, $\overline{DA}$), and ($\overline{DA}$, $\overline{AB}$).
- Opposite Sides: Two sides of a quadrilateral are called opposite if they do not share a common vertex. They are located across from each other. In quadrilateral ABCD, the pairs of opposite sides are: ($\overline{AB}$, $\overline{CD}$) and ($\overline{BC}$, $\overline{DA}$).
- Adjacent Angles: Two angles of a quadrilateral are called adjacent if they share a common side as one of their arms (and their vertices are the endpoints of that side). In quadrilateral ABCD, the pairs of adjacent angles are: ($\angle A$, $\angle B$), ($\angle B$, $\angle C$), ($\angle C$, $\angle D$), and ($\angle D$, $\angle A$).
- Opposite Angles: Two angles of a quadrilateral are called opposite if they are not adjacent. They are located across from each other at opposite vertices. In quadrilateral ABCD, the pairs of opposite angles are: ($\angle A$, $\angle C$) and ($\angle B$, $\angle D$).
Convex and Concave Quadrilaterals
Quadrilaterals can be broadly classified into two types based on their interior angles and the position of their diagonals:
- Convex Quadrilateral:
A quadrilateral is said to be convex if all its interior angles are less than $180^\circ$. Equivalently, a quadrilateral is convex if for any side, the entire quadrilateral lies on the same side of the line containing that side. Another way to identify a convex quadrilateral is that both of its diagonals lie entirely within the interior of the quadrilateral.
- Concave Quadrilateral:
A quadrilateral is said to be concave if at least one of its interior angles is a reflex angle (greater than $180^\circ$). In a concave quadrilateral, at least one of the diagonals lies either partially or wholly outside the interior of the quadrilateral. A concave quadrilateral is sometimes also called a reentrant 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.

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.

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.

In the figure, ABCD is an isosceles trapezium with $AB \parallel DC$ and $AD = BC$.
Properties of an Isosceles Trapezium:
- The base angles are equal. That is, the angles formed at each base are equal. $\angle DAB = \angle CBA$ and $\angle ADC = \angle BCD$.
- The diagonals are equal in length. $\overline{AC}$ and $\overline{BD}$ have the same length ($AC = BD$).
- Opposite angles are supplementary (sum up to $180^\circ$). Note that this is only true if the definition requires exactly one pair of parallel sides.
2. Kite
Definition: A Kite is a quadrilateral that has two distinct pairs of equal-length adjacent sides.

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:
- The diagonals are perpendicular to each other. $\overline{AC} \perp \overline{BD}$.
- One diagonal (the one connecting the vertices where the unequal sides meet, $\overline{AC}$ in the figure) is the perpendicular bisector of the other diagonal ($\overline{BD}$). Thus, this diagonal bisects the other diagonal ($BO = OD$).
- One diagonal (the one connecting the vertices where the unequal sides meet, $\overline{AC}$) bisects the pair of opposite angles where the equal adjacent sides meet ($\angle DAB$ is bisected by $\overline{AC}$, and $\angle BCD$ is bisected by $\overline{AC}$).
- One pair of opposite angles are equal (the angles between the unequal sides, $\angle ABC$ and $\angle ADC$). Note that the angles where the equal adjacent sides meet ($\angle DAB$ and $\angle BCD$) are generally not equal unless the kite is also a rhombus or square.
3. Parallelogram
Definition: A Parallelogram is a quadrilateral in which both pairs of opposite sides are parallel.

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.

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:
- All four interior angles are right angles ($90^\circ$).
- The diagonals are equal in length ($AC = BD$).
- Opposite sides are equal and parallel (inherited from parallelogram).
- Diagonals bisect each other (inherited from parallelogram).
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.

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:
- All four sides are equal in length ($AB = BC = CD = DA$).
- The diagonals bisect each other at right angles ($\overline{AC} \perp \overline{BD}$ and they intersect at their midpoints).
- The diagonals bisect the interior angles at the vertices. For example, the diagonal $\overline{AC}$ bisects $\angle A$ and $\angle C$, and the diagonal $\overline{BD}$ bisects $\angle B$ and $\angle D$.
- Opposite angles are equal (inherited from parallelogram).
- Opposite sides are parallel (inherited from parallelogram).
3. Square
Definition: A Square is the most special type of parallelogram. It can be defined in several equivalent ways:
- A parallelogram with all four sides equal and one right angle.
- A rectangle with adjacent sides equal.
- A rhombus with one right angle.
- A quadrilateral with 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:
- All four sides are equal.
- All four interior angles are right angles ($90^\circ$).
- The diagonals are equal in length.
- The diagonals bisect each other at right angles.
- The diagonals bisect the interior angles at the vertices, resulting in $45^\circ$ angles ($\angle BAC = \angle CAD = \angle BCA = \angle ACD = 45^\circ$, etc.).
- Opposite sides are parallel.
- Opposite angles are equal (which is $90^\circ$).
Hierarchy and Relationships among Quadrilaterals
The relationship between these special quadrilaterals can be understood as a hierarchy:
- Every Square is both a Rectangle and a Rhombus because it satisfies the definitions of both.
- Every Rectangle and every Rhombus is a Parallelogram because they satisfy the definition of a parallelogram.
- Every Parallelogram, Rectangle, Rhombus, Square, Trapezium (using the 'at least one pair' definition), and Kite is a Quadrilateral because they are all closed figures with four sides.
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.

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:
- If both pairs of opposite sides of a quadrilateral are equal in length, then it is a parallelogram.
- If both pairs of opposite angles of a quadrilateral are equal, then it is a parallelogram.
- If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.
- If one pair of opposite sides of a quadrilateral is both equal in length and parallel, then it is a parallelogram.
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.

Inherited Parallelogram Properties:
- Opposite sides are equal and parallel ($\overline{AB} \parallel \overline{DC}$, $\overline{AD} \parallel \overline{BC}$, $AB = DC$, $AD = BC$).
- Opposite angles are equal ($\angle A = \angle C$, $\angle B = \angle D$). Since all angles are $90^\circ$, this is trivially true ($90^\circ = 90^\circ$).
- Consecutive angles are supplementary ($\angle A + \angle B = 90^\circ + 90^\circ = 180^\circ$, etc.).
- Diagonals bisect each other (if diagonals $\overline{AC}$ and $\overline{BD}$ intersect at O, then $AO = OC$ and $BO = OD$).
Specific Properties of a Rectangle:
- Angle Property: All four interior angles are right angles ($\angle A = \angle B = \angle C = \angle D = 90^\circ$). This is part of its definition.
- Diagonal Property: The diagonals of a rectangle are equal in length ($AC = BD$).
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.

Inherited Parallelogram Properties:
- Opposite sides are parallel ($\overline{AB} \parallel \overline{DC}$, $\overline{AD} \parallel \overline{BC}$).
- Opposite angles are equal ($\angle A = \angle C$, $\angle B = \angle D$).
- Consecutive angles are supplementary.
- Diagonals bisect each other (if diagonals $\overline{AC}$ and $\overline{BD}$ intersect at O, then $AO = OC$ and $BO = OD$).
Specific Properties of a Rhombus:
- Side Property: All four sides are equal in length ($AB = BC = CD = DA$). This is part of its definition.
- Diagonal Property 1: The diagonals of a rhombus bisect each other at right angles ($\overline{AC} \perp \overline{BD}$, meaning $\angle AOB = \angle BOC = \angle COD = \angle DOA = 90^\circ$).
- Diagonal Property 2: The diagonals of a rhombus bisect the interior angles at the vertices (e.g., $\overline{AC}$ bisects $\angle A$ and $\angle C$, $\overline{BD}$ bisects $\angle B$ and $\angle D$).
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.

Properties of a Square:
- Side Property: All four sides are equal (inherited from rhombus).
- Angle Property: All four interior angles are right angles (inherited from rectangle).
- Diagonal Property 1: Diagonals bisect each other (inherited from parallelogram).
- Diagonal Property 2: Diagonals are equal in length (inherited from rectangle).
- Diagonal Property 3: Diagonals bisect each other at right angles (inherited from rhombus).
- Diagonal Property 4: Diagonals bisect the interior angles at the vertices (inherited from rhombus). Since all vertex angles are $90^\circ$, each diagonal bisects the angles into two $45^\circ$ angles (e.g., $\angle BAC = \angle BCA = \angle CAD = \angle ACD = 45^\circ$).
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.

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:
- Parallel Sides Property: It has at least one pair of opposite sides that are parallel.
- Consecutive Angles Property: The angles between a pair of parallel sides and a leg are supplementary. If $AB \parallel DC$, then:
$\angle D + \angle A = 180^\circ$
(Consecutive interior angles between parallel lines AD and BC cut by transversal AB) $\times$
This statement is incorrect. The correct statement refers to angles between parallel sides and a leg.
$\angle A + \angle D = 180^\circ$
(Consecutive interior angles between parallel lines AB and DC cut by transversal AD)
$\angle B + \angle C = 180^\circ$
(Consecutive interior angles between parallel lines AB and DC cut by transversal BC)
- Median (or Midsegment) Property: The line segment joining the midpoints of the non-parallel sides of a trapezium is called the median or midsegment. This median is parallel to the bases, and its length is half the sum of the lengths of the bases. If M is the midpoint of $\overline{AD}$ and N is the midpoint of $\overline{BC}$, then $MN \parallel AB \parallel DC$ and $MN = \frac{1}{2}(AB + DC)$.
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 $AD = BC$.

In addition to the properties of a general trapezium, an isosceles trapezium has these specific properties:
- Legs Property: The non-parallel sides are equal in length ($AD = BC$). This is part of its definition.
- Base Angles Property: The angles at each base are equal. That is, the angles formed by a base and the two legs are equal.
$\angle D = \angle C$
(Base angles at base DC)
$\angle A = \angle B$
(Base angles at base AB)
- Diagonal Property: The diagonals are equal in length ($AC = BD$).
- Opposite Angles Property: Opposite angles are supplementary ($\angle A + \angle C = 180^\circ$ and $\angle B + \angle D = 180^\circ$). This follows from the base angle property and the supplementary property of angles between parallel sides and a leg. For example, $\angle A + \angle D = 180^\circ$ (general trapezium property). Since $\angle D = \angle C$ (isosceles trapezium property), substituting gives $\angle A + \angle C = 180^\circ$. Similarly for $\angle B + \angle D$.
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.

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:
- Side Property: It has two distinct pairs of adjacent sides that are equal.
- Diagonal Property 1: The diagonals are perpendicular to each other ($AC \perp BD$).
- Diagonal Property 2: One diagonal (the one connecting the vertices between equal sides, $\overline{AC}$) is the perpendicular bisector of the other diagonal ($\overline{BD}$). This means that the diagonal $\overline{AC}$ cuts the diagonal $\overline{BD}$ into two equal parts ($BO = OD$) at a right angle. (Note: The diagonal $\overline{BD}$ does NOT necessarily bisect $\overline{AC}$).
- Diagonal Property 3: One diagonal (the one connecting the vertices between equal sides, $\overline{AC}$) bisects the angles at the two vertices it connects ($\angle BAC = \angle DAC$ and $\angle BCA = \angle DCA$).
- Angle Property: One pair of opposite angles (the angles between the unequal sides, $\angle ABC$ and $\angle ADC$) are equal. Note that the angles at the vertices between the equal sides ($\angle BAD$ and $\angle BCD$) are generally not equal, unless the kite is also a rhombus (and hence a square).
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.