Theorems Related to Quadrilaterals: If all the points are collinear, we obtain a line segment, if three out of four points are collinear, we get a triangle, and if no, three points out of four are collinear, we obtain a closed figure with four sides. Such a figure formed by joining four points in order is called a quadrilateral. A quadrilateral has four sides, four angles and four vertices.

This article will study the definition of quadrilateral, its types, and different theorems related to quadrilaterals and solve some example problems.

Quadrilaterals

The word quad means four, and the word lateral means sides. Thus, a plane figure bounded by four line segments \(AB,\,BC,\,CD\) and \(DA\) is called a quadrilateral and is written as quad. \(ABCD\). The points \(A,\,B,\,C,\,D\) are called vertices. The four-line segments \(AB,\,BC,\,CD\) and \(DA\) are the four sides, and the four angles \(\angle A,\,\angle B,\,\angle C,\,\angle D\) are the four angles of the quad. \(ABCD\) Two line segments \(AC\) and \(BD\) are called the diagonals of the quad. \(ABCD\)

Below are some of the definitions related to quadrilateral:

Adjacent sides: Two sides of the quadrilateral are consecutive or adjacent sides if they have a common point or vertex.

Opposite Sides: Two sides of a quadrilateral are opposite sides if they have no common end-points or vertex.

Consecutive Angles: The consecutive angles of a quadrilateral are two angles that include a side in their intersection.

Opposite Angles: Two angles of a quadrilateral are said to be opposite angles if they do not have a common arm.

Angle Sum Property of a Quadrilateral

Theorem: The sum of the four angles of a quadrilateral is \({360^ \circ }\)

Given: Quadrilateral \(ABCD\)

To prove \(\angle A\, + \,\angle B + \angle C + \angle D = {360^ \circ }\)

Construction: Join \(AC\)

Proof: In \(\Delta ABC\), we have \(\angle 1 + \angle 4 + \angle 6 = {180^ \circ }………(i)\)

In \(\Delta ACD,\) we have \(\angle 2 + \angle 3 + \angle 5 = {180^ \circ }………(ii)\)

Adding (i) and (ii), we get,

\((\angle 1 + \angle 2) + (\angle 3 + \angle 4) + (\angle 5 + \angle 6) = {180^ \circ } + {180^ \circ }\)

\( \Rightarrow \angle A + \angle B + \angle C + \angle D = {360^ \circ }\)

Types of Quadrilateral

Let us learn about several types of quadrilateral.

Trapezium: A quadrilateral having one pair of parallel sides is called a trapezium.

Isosceles Trapezium: A trapezium is said to be isosceles if its non-parallel sides are equal.

Parallelogram: A quadrilateral is a parallelogram if both pairs of opposite sides are parallel.

Rhombus: A parallelogram having all sides are equal is called a rhombus.

Rectangle: A parallelogram whose each angle is a right angle is called a rectangle.

Square: A square is a rectangle with a pair of adjacent sides equal.

Kite: A quadrilateral is a kite with two pairs of equal adjacent sides and unequal opposite sides.

Properties of a Parallelogram

In this section, we shall prove some theorems of a parallelogram.

Theorem 1: A diagonal of a parallelogram divides it into two congruent triangles.

Given: A parallelogram \(ABCD\)

To prove: A diagonal \(AC\) of parallelogram \(ABCD\) divides it into congruent triangles \(ABC\) and \(CDA,\) i.e. \(\Delta ABC \cong \Delta CDA\)

Construction: Join \(AC\)

Proof: Since \(ABCD\) is a parallelogram. Therefore, \(AB\parallel DC\) and \(AD\parallel BC\)

Now, \(AD\parallel BC\) and transversal \(AC\) intersects them at \(A\)  and \(C\) respectively.

Therefore, \(\angle DAC\, = \,\angle BCA………(i)\) [Alternate interior angles]

Again \(AB\parallel DC\) and transversal \(AC\) intersects them at \(A\) and \(C\) respectively.

Therefore, \(\angle BAC\, = \,\angle DCA…….(ii)\) [Alternate interior angles]

Now, in triangles \(ABC\) and \(CDA\) we have

\(\angle DAC = \angle BCA\) [From (i)]

\(AC = AC\) [Common side]

\(\angle BAC = \angle DCA\) [From (ii)]

So, by ASA congruence criterion, we have

\(\Delta ABC \cong \Delta CDA\)

Theorem 2: In a parallelogram, opposite sides are equal.

Given: A parallelogram \(ABCD\)

To prove: \(AB\, = CD\) and \(DA\, = BC\)

Construction: Join \(AC\)

Proof: Since \(ABCD\) is a parallelogram. Therefore, \(AB\parallel DC\) and \(AD\parallel BC\)

Now, \(AD\parallel BC\) and transversal \(AC\) intersects them at \(A\) and \(C\) respectively.

Therefore, \(\angle DAC = \angle BCA\)……(i) (Alternate interior angles)
Again \(AB\parallel DC\) and transversal \(AC\) intersects them at \(A\) and \(C\) respectively.
Therefore, \(\angle BAC = \angle DCA\)……(ii) (Alternate interior angles)
Now, in triangles \(ABC\) and \(CDA,\) we have:
\(\angle DAC = \angle BCA\) (From (i))
\(AC = AC\) (Common side)
\(\angle BAC = \angle DCA\) (From (ii))
Now, in triangles \(ADC\) an\(CBA,\) we have:
\(\angle DAC = \angle BCA\) (From (i))
\(AC = AC\) (Common side)
\(\angle BAC = \angle DCA\) (From (ii))

So, by ASA criterion of congruence,

\(\Delta ADC \cong \Delta CBA\)
\( \Rightarrow AD = CB\) and \(DC = BA\) (Corresponding parts of congruent triangles are equal)

Theorem 3: The opposite angles of a parallelogram are equal.

Given: A parallelogram \(ABCD\)

To prove: \(\angle A=\angle C\) and \(\angle B=\angle D\)

Proof: Since \(ABCD\) is a parallelogram. Therefore, \(AB\parallel DC\) and \(AD\parallel BC\)

Now, \(AB\parallel DC\) and transversal \(AD\) intersects them at \(A\) and \(D\) respectively.

\(\angle A + \angle D = {180^ \circ } \ldots \ldots (i)\) [ Sum of consecutive interior angles is \({180^ \circ }\)

Again \(AD\parallel BC\) and \(DC\) intersects them at \(D\) and \(C\) respectively.

\(\angle D + \angle C = {180^ \circ } \ldots \ldots (ii)\) [ Sum of consecutive interior angles is \({180^ \circ }\)

From (i) and (ii), we get,

\(\angle A+\angle D=\angle D+\angle C\)

\(\Rightarrow \angle A=\angle C\)

Similarly, \(\angle B=\angle D\)

So, \(\angle A=\angle C\) and \(\angle B=\angle D\)

Theorem 4: The diagonals of a parallelogram bisect each other

Given:  A parallelogram \(ABCD\) such that its diagonal \(AC\) and \(BD\) intersect at \(O.\)

To prove: \(OA=OC\) and \(OB=OD\)

Proof: Since \(ABCD\) is a parallelogram. Therefore, \(AB\parallel DC\) and \(AD\parallel BC\)

Now, \(AB\parallel DC\) and transversal \(AC\) intersects them at \(A\) and \(C\) respectively.

Therefore, \(\angle B A C=\angle D C A\) [Alternate interior angles are equal]

\(\Rightarrow \angle B A O=\angle D C O….(i)\)

Again, \(AD\parallel BC\) and \(BD\) intersects them at \(B\) and \(D\) respectively.

Therefore, \(\angle A B D=\angle C D B\)

\(\Rightarrow \angle A B O=\angle C D O ……(ii)\) [Alternate interior angles are equal]

Now in triangles \(AOB\) and \(COD,\)

\(\angle B A O=\angle D C O\) [From (i)]

\(AB=CD\) [Opposite sides of a parallelogram] and,

\(\angle A B O=\angle C D O\) [From (ii)]

So, by \(ASA\) congruence criterion,

\(\Delta AOB \cong \Delta COD\)

\(\Rightarrow O A=O C\) and \(OB=OD\)

Theorem 5: In a parallelogram, the bisectors of any two consecutive angles intersect at a right angle.

Given: A parallelogram \(ABCD\) such that the bisectors of consecutive angles \(A\) and \(B\) intersects at \(P.\)

To prove \(\angle A P B=90^{\circ}\)

Proof: Since \(ABCD\) is a parallelogram. Therefore, \(AD∥BC\)

Now, \(AD∥BC\) and transversal \(AB\) intersects them.

Therefore, \(\angle A+\angle B=180^{\circ}\)  [Sum of consecutive interior angles is \(\left.180^{\circ}\right]\)

\(\Rightarrow \frac{1}{2} \angle A+\frac{1}{2} \angle B=90^{\circ}\)

Since \(AP\) is the bisector of \(\angle A\) and \(BP\) is the bisector of \(\angle B\) So, \(\angle 1=\frac{1}{2} \angle A\) and \(\angle 2=\frac{1}{2} \angle B\)

\(\Rightarrow \angle 1+\angle 2=90^{\circ} \ldots \ldots (i)\)

In \(\Delta APB\) we have,

\(\angle 1+\angle A P B+\angle 2=180^{\circ}\)

\(\Rightarrow \angle A P B+90^{\circ}=180^{\circ}\)[From (i)]

\(\Rightarrow \angle A P B=90^{\circ}\)

Theorem 6: If the diagonal of a parallelogram bisects one of the angles of the parallelogram, it also bisects the second angle. Also, prove that it is a rhombus.

Given: A parallelogram \(ABCD\) in which diagonal \(AC\) bisects \(\angle A\)

To prove: \(AC\) bisects \(\angle C\)

Proof: Since \(ABCD\) is a parallelogram. Therefore, \(AB∥DC\)

Now, \(AB∥DC\) and transversal \(AC\) intersects them.

Therefore, \(\angle 1=\angle 3\) ……(i) [Alternate interior angles]

Again, \(AD∥BC\) and transversal \(AC\) intersects them.

Therefore, \(\angle 2=\angle 4 \ldots \ldots(i i)\)[Alternate interior angles]

But, it is given that \(AC\) is the bisector of \(\angle A\) Therefore, 

\(\angle 1=\angle 2……(iii)\)

From (i), (ii) and (iii), we get:

\(\angle 3=\angle 4……(iv)\)

Hence, \(AC\) bisects \(\angle C\)

From (ii) and (iii), we have:

\(\angle 1=\angle 4\)

\(\Rightarrow B C=A B\) [Angles opposite to equal sides are equal]

But, \(AB=DC\) and \(BC=AD\) \([ABCD\) is a parallelogram]

So, \(AB=BC=CD=DA\)

Hence, \(ABCD\) is a rhombus.

Theorem 7: The angle bisectors of a parallelogram form a rectangle.

Given: A parallelogram \(ABCD\) in which bisectors of angles \(A, B, C, D\) intersects at \(P, Q, R, S\) to form a quadrilateral \(PQRS\)

Proof: Since \(ABCD\) is a parallelogram. Therefore, \(AD∥BC\)

Now, \(AD∥BC\) and transversal \(AB\) intersects them.

Therefore, \(\angle A+\angle B=180^{\circ}\)  [Sum of consecutive interior angles is \(\left.180^{\circ}\right]\)

\(\Rightarrow \frac{1}{2} \angle A+\frac{1}{2} \angle B=90^{\circ}\)

\(\Rightarrow \angle B A S+\angle A B S=90^{\circ} \ldots \ldots(i)\) [\(AS\) and \(BS\) are bisectors of \(\angle A\) and \(\angle B\) respectively]

But in \(\Delta ABS\) we have,

\(\angle B A S+\angle A B S+\angle A S B=180^{\circ}\)

\(\Rightarrow 90^{\circ}+\angle A S B=180^{\circ}\)

\(\Rightarrow \angle A S B=90^{\circ}\)

\(\Rightarrow \angle R S P=90^{\circ}[\angle A S B\) and \(\angle R S P\) are vertically opposite angles]

Similarly, we can prove that \(\angle S R Q=90^{\circ}, \angle R Q P=90^{\circ}\) and \(\angle S P Q=90^{\circ}\)

Hence, \(PQRS\) is a rectangle.

Conditions for a Quadrilateral to be a Parallelogram

In this section, we will study some conditions which a quadrilateral must satisfy to become a parallelogram.

Theorem 1: A quadrilateral is a parallelogram if its opposite sides are equal.

Given: A quadrilateral \(ABCD\) in which \(AB=CD\) and \(BC=DA\)

To prove: \(ABCD\) is a parallelogram

Construction: Join \(AC\)

Proof: In triangles \(ACB\) and \(CAD,\) we have, 

\(AC=CA\) [Common side]

\(CB=AD\) [Given]

\(AB=CD\) [Given]

So, by the \(SSS\) criterion of congruence, we have, 

\(\Delta ACB \cong \Delta CAD\)

\(\Rightarrow \angle C A B=\angle A C D ……(i)\) [Corresponding parts of congruent triangles are equal] and,

\(\angle A C B=\angle C A D\)

Now, line \(AC\) intersects \(AB\) and \(DC\) at \(A\) and \(C,\) such that:

\(\angle C A B=\angle A C D\) [From (i)]

i.e., alternate interior angles are equal.

Therefore, \(AB∥DC\)

Similarly, line \(AC\) intersects \(BC\) and \(AD\) at \(C\) and \(A\), such that,

\(\angle A C B=\angle C A D\)

i.e., alternate interior angles are equal.

Therefore, \(BC∥AD\)

So, \(AB∥DC\) and \(BC∥AD\)

\(\Rightarrow A B C D\) is a parallelogram.

Theorem 2: A quadrilateral is a parallelogram if its opposite angles are equal.

Given: A quadrilateral \(ABCD\) in which \(\angle A=\angle C\) and \(\angle B=\angle D\)

To prove: \(ABCD\) is a parallelogram

Proof: In a quadrilateral \(ABCD,\) we have,

\(\angle A=\angle C_{. . .}…(i)\)  [Given]

\(\angle B=\angle D \ldots \ldots . .(\mathrm{ii})\)[Given]

Adding (i) and (ii), we get

\(\angle A+\angle B=\angle C+\angle D ……..(iii)\)

Since the sum of the angles of a quadrilateral is \(360^{\circ} .\)

Therefore, \(\angle A+\angle B+\angle C+\angle D=360^{\circ} \ldots . .(iv)\)

\(\Rightarrow \angle A+\angle B+\angle A+\angle B=360^{\circ}\) [From (iii)]

\(\Rightarrow 2(\angle A+\angle B)=360^{\circ}\)

\(\Rightarrow(\angle A+\angle B)=180^{\circ}\)

\(\Rightarrow \angle A+\angle B=\angle C+\angle D=180^{\circ}\)

Now, line \(AB\) intersects \(AD\) and \(BC\) at \(A\) and \(B\) respectively, such that,

\(\angle A+\angle B=180^{\circ}\)

i.e. the sum of consecutive interior angles is \(180^{\circ}\)

\( \Rightarrow AD\parallel BC\)

Again, \(\angle A+\angle B=180^{\circ} \Rightarrow \angle C+\angle D=180^{\circ}\)

Now, line \(BC\) intersects \(AB\) and \(DC\) at \(A\) and \(C\) respectively such that,

\(\angle B+\angle C=180^{\circ}\)

i.e. the sum of consecutive interior angles is \(180^{\circ}\)

\( \Rightarrow AB\parallel DC\)

So, \(AB∥DC\) and \(AD∥BC\)

\(\Rightarrow A B C D\) is a parallelogram.

Theorem 3: If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram.

Given: A quadrilateral \(ABCD\) in which the diagonals \(AC\) and \(BD\) intersects at \(O\) such that \(AO=OC\) and \(BO=OD\)

To prove: Quadrilateral \(ABCD\) is a parallelogram

Proof: In triangles \(AOD\) and \(COB,\) we have,

\(AO=OC\) [Given]

\(OD=OB\) [Given]

\(\angle A O D=\angle C O B\) [Vertically opposite angles]

So, by SAS criterion of congruence, we have,

\(\Delta AOD \cong \Delta COB\)

\(\Rightarrow \angle O A D=\angle O C B\) [Corresponding parts of congruent triangles are equal] ……(i)

Now, line \(AC\) intersects \(AD\) and \(BC\) at \(A\) and \(C\) respectively such that,

\(\angle O A D=\angle O C B\) [From (i)]

i.e., alternate interior angles are equal.

Therefore, \(AD∥BC\)

Similarly, \(AB∥DC\)

Hence, \(ABCD\) is a parallelogram.

Theorem 4: A quadrilateral is a parallelogram if one pair of opposite sides are equal and parallel.

Given: A quadrilateral \(ABCD\) in which \(AB=CD\) and \(AB∥CD\)

To prove: \(ABCD\) is a parallelogram

Construction: Join \(AC\)

Proof: In triangles \(ABC\) and \(CDA,\) we have,

\(AB=DC\) [Given]

\(AC=AC\) [Common sides] and,

\(\angle B A C=\angle D C A\) [Alternative angles are equal]

So, by SAS criterion of congruence, we have,

\(\Delta ABC \cong \Delta CDA\)

\(\Rightarrow \angle B C A=\angle D A C\) [CPCT]

Thus, line \(AC\) intersects \(AB\) and \(DC\) at \(A\) and \(C\) respectively such that,

\(\angle D A C=\angle B C A\)

i.e., alternate angles are equal

\( \Rightarrow AD\parallel BC\)

So, \(ABCD\) is a parallelogram.

Theorems on Cyclic Quadrilaterals

When a quadrilateral is inscribed in a circle, i.e., the quadrilateral’s vertices lie on the circumference of a circle. The quadrilateral is called a cyclic quadrilateral.

The points, which lie on the circumference of the same circle, are called concyclic points.

Theorem 1: The opposite angles of a cyclic quadrilateral (quadrilateral inscribed in a circle) are supplementary.

To Prove: \(\angle A B C+\angle A D C=180^{\circ}\) and \(\angle B A D+\angle B C D=180^{\circ}\)

Construction: Join \(OA\) and \(OC\)

Proof: Arc \(ABC\) subtends angle \(AOC\) at the centre and angle \(ADC\) at point \(D\) of the remaining circumference.

Therefore, \(\angle AOC = 2\angle ADC\)[Angle at the centre is twice the angle at remaining circumference]

\( \Rightarrow \angle ADC = \frac{1}{2}\angle AOC\)

Similarly, \(\angle ABC = \frac{1}{2}\) reflex \(\angle AOC\)

Now, \(\angle ABC + \angle ADC = \frac{1}{2}({\rm{ reflex }}\angle AOC + \angle AOC)\)

\(\Rightarrow \angle A B C+\angle A D C=\frac{1}{2} \times 360^{\circ}=180^{\circ}\)

Similarly \(\angle B A D+\angle B C D=180^{\circ}\)

Theorem 2: The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.

Given: A cyclic quadrilateral \(ABCD\) whose side \(AB\) is produced to a point \(E\)

To prove: Ext. \(\angle CBE = \angle ADC\)

Proof:

\(\angle A B C+\angle C B E=180^{\circ}\)

\(\angle A B C+\angle A D C=180^{\circ}\) [Opposite angles of a cyclic quadrilateral are supplementary]

Therefore, \(\angle ABC + \angle CBE = \angle ABC + \angle ADC\)

\( \Rightarrow \angle CBE = \angle ADC\)

Solved Examples – Theorems Related to Quadrilaterals

Q.1: The diagonals of a rectangle \(ABCD\) meet at \(O\). If \(\angle B O C=44^{\circ}\) find \(\angle OAD.\)

Ans: We have \(\angle B O C+\angle B O A=180^{\circ}\)
\(\Rightarrow 44^{\circ}+\angle B O A=180^{\circ}\)
\(\Rightarrow \angle B O A=136^{\circ}\)
Since diagonals of a rectangle are equal and they bisect each other. Therefore, in \(\Delta OAB\) we have, 
\(OA=OB\) [Angles opposite to equal sides are equal]
\(\Rightarrow \angle 1=\angle 2\)
Now, in \(\Delta OAB\) we have, 
\(\angle 1+\angle 2+\angle B O A=180^{\circ}\)
\(\Rightarrow 2 \angle 1+136^{\circ}=180^{\circ}\)
\(\Rightarrow 2 \angle 1=44^{\circ}\)
\(\Rightarrow \angle 1=22^{\circ}\)
Since each angle of a rectangle is a right angle.
Therefore, \(\angle B A D=90^{\circ}\)
\(\Rightarrow \angle 1+\angle 3=90^{\circ}\)
\(\Rightarrow 22^{\circ}+\angle 3=90^{\circ}\)
\(\Rightarrow \angle 3=68^{\circ}\)
\(\Rightarrow \angle O A D=68^{\circ}\)

Q.2: \(PQRS\) is a square. Determine \(\angle SRP{\rm{. }}\)

Ans: Since \(PQRS\) is a square.
Therefore, \(PS\, = \,SR\) and \(\angle P S R=90^{\circ}\)
Now, in \(\Delta PSR\) we have,
\(PS = SR\)
\( \Rightarrow \angle 1 = \angle 2\) (Angles opposite to equal sides are equal)
But, \(\angle 1 + \angle 2 + \angle PSR = {180^{\rm{o}}}\)
\( \Rightarrow 2\angle 1 + {90^{\rm{o}}} = {180^{\rm{o}}}\)
\( \Rightarrow 2\angle 1 = {90^{\rm{o}}}\)
\( \Rightarrow \angle 1 = {45^{\rm{o}}}\)
\( \Rightarrow \angle SRP = {45^{\rm{o}}}\)

Q.3: \(ABCD\) is a rhombus with \(\angle ABC = {56^{\rm{o}}}.\) Determine \(\angle ACD\)

Ans: \(ABCD\) is a rhombus
\( \Rightarrow ABCD\) is a parallelogram
\( \Rightarrow \angle ABC = \angle ADC\)
\( \Rightarrow \angle ADC = {56^{\rm{o}}}\)
\( \Rightarrow \angle ODC = {28^{\rm{o}}}\,\left[ {\angle ODC = \frac{1}{2}\angle ADC} \right]\)
Now, in \(\Delta OCD,\) we have,
\(\angle OCD + \angle ODC + \angle COD = {180^{\rm{o}}}\)
\( \Rightarrow \angle OCD + {28^{\rm{o}}} + {90^{\rm{o}}} = {180^{\rm{o}}}\)
\( \Rightarrow \angle OCD = {62^{\rm{o}}}\)
\( \Rightarrow \angle ACD = {62^{\rm{o}}}\)

Q.4: In a parallelogram, \(ABCD\) prove that sum of any two consecutive angles is \({180^ \circ }\).

Ans: Since \(ABCD\) is a parallelogram. Therefore, \(AD\parallel BC\)
Now, \(AD\parallel BC\) and transversal \(AB\) intersects them at \(A\) and \(B\) respectively.
Therefore, \(\angle A + \angle B = {180^{\rm{o}}}\) (Sum of the interior angles on teh same side of transversal is \({180^{\rm{o}}}\))
Similarly, we can prove that \(\angle B + \angle C = {180^{\rm{o}}},\,\angle C + \angle D = {180^{\rm{o}}}\) and \(\angle D + \angle A = {180^{\rm{o}}}.\)

Q.5: Show that the diagonals of a square are equal and bisect each other at right angles.
Ans: Let \(ABCD\) be a square such that its diagonals \(AC\) and \(BD\)  intersect at \(O\) We have to prove that \(AC = BD\) and \(AC\) and \(BD\) bisect each other at the right angle.

In triangles \(BAD\) and \(CDA\) we have,
\(BA\, = \,CD\)
\(\angle BAD = \angle CDA\) (Each equal to \({90^ \circ }\))
\(AD = DA\) (Common)
So, by SAS congruence criterion, we have,
\(\Delta BAD \cong \Delta CDA\)
\( \Rightarrow BD = CA\)
Since every square is a parallelogram and diagonals of a parallelogram bisect each other.
Therefore, \(OA = OC\) and \(OB = OD\)
Now, consider triangles \(AOB\) and \(COB\) In these triangles, we have,
\(AB = CB\) (\(ABCD\) is a square)
\(OD = BD\) (Common) and,
\(OA = OC\)
So, by the SSS congruence criterion, we have,
\(\Delta AOB \cong \Delta COB\)
\( \Rightarrow \angle AOB = \angle COB\)
But, \(\angle AOB + \angle COB = {180^{\rm{o}}}\)
Therefore, \(\angle AOB = \angle COB = {90^{\rm{o}}}\)
Hence, the diagonals of square \(ABCD\) are equal and bisect each other at right angles.

Summary

In this article, we have learnt the definition of quadrilaterals and different types of quadrilaterals. Also, we have proved theorems on cyclic quadrilaterals, theorems for a quadrilateral to be a parallelogram and theorems on properties of a parallelogram and solved some example problems.

Frequently Asked Questions (FAQs) – Theorems Related to Quadrilaterals

Q.1. What are the different types of quadrilaterals?
Ans: The different types of quadrilaterals are:
Parallelogram, square, rectangle, trapezium, rhombus and kite.

Q.2. Is a kite quadrilateral?
Ans: A kite is a quadrilateral whose four sides can be grouped into two pairs of equal-length sides adjacent to each other.

Q.3. Is a rhombus square?
Ans: A rhombus is a quadrilateral with four equal sides and opposite sides parallel to each other. All squares are rhombuses, but not all rhombuses are squares.

Q.4. What is a cyclic quadrilateral?
Ans: When a quadrilateral is inscribed in a circle, i.e., the quadrilateral’s vertices lie on the circumference of a circle. The quadrilateral is called a cyclic quadrilateral.

Q.5. What are consecutive angles in a quadrilateral?
Ans: The consecutive angles of a quadrilateral are two angles that include a side in their intersection.

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NCERT Solutions For Chapter: Quadrilaterals

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