CBSE Notes For Class 9 Maths Chapter 8 Quadrilaterals 

1.0Introduction to Quadrilaterals 

The word "Quadrilateral" originates from two Latin words: "quadri", meaning four, and "latus" referring to the side. This makes it a 4-sided figure in geometry. These can be of different shapes and sizes: 

2.0Download CBSE Notes for Class 9 Maths Chapter 8 Quadrilaterals - Free PDF

Students can now download free, well-structured PDF notes for CBSE Class 9 Maths Chapter 8 Quadrilaterals, designed to help simplify and strengthen their understanding of this important geometry chapter.

3.0CBSE Class 9 Maths Chapter 8 Quadrilaterals - Revision Notes

What is a Parallelogram? 

A parallelogram is a type of quadrilaterals which have opposite sides equal & parallel to each other. Some properties of parallelograms are as follows: 

Properties of A Parallelogram

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

To prove: ABC ≅ CDA 

Given: ABCD is a parallelogram, which means CD is parallel to AB

Proof: In ABC and CDA

∠BCA = ∠DAC (Alternate interior angle) 

∠BAC = ∠DCA (Alternate interior angle) 

AC = AC (Common) 

ABC ≅  CDA (ASA) 

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

To Prove: AB = CD, and AD = BC

Proof: As mentioned above, 

ABC ≅ CDA

Hence,

AB = CD (CPCT = Congruent Parts of Congruent Triangles) 

AD = BC (CPCT)

Theorem-3: If each pair of opposite sides of any quadrilateral is equal, then it is a parallelogram. 

To Prove: ABCD is a parallelogram.

Given: AB = CD, and BC = DA

Proof: In ABD and CDB

AB = CD (Given)

BC = DA (Given)

BD = BD (Common) 

 ABD ≅  CDB (SSS)

Hence, ∠ ADB = ∠ DBC (CPCT)

And ∠ ABD = ∠CDB (CPCT)

AB is parallel to CD, and BC is parallel to AD (Inverse of Alternate interior angle theorem) 

Theorem-4: In a parallelogram, Opposite angles are equal. 

To Prove: ∠DAB = ∠DCB and ∠ADC = ∠ABC

Proof: From the above equation, 

ABD ≅ CDB

Hence, ∠DAB = ∠DCB (CPCT) 

Similarly, ADC ≅ CBA 

∠ADC = ∠ABC

Theorem-5: A given quadrilateral is a parallelogram if each pair of opposite angles is equal. 

To Prove: ABCD is a parallelogram.

Proof: aA = aC and aB = aD 

Solution: By angle sum property of a quadrilateral, 

∠A + ∠B + ∠C +∠D = 360 

∠A + ∠B + ∠A + ∠B = 360 

∠A + ∠B = 180 

In maths, the sum of the adjacent angles of a parallelogram is always equal to 180 degrees. Hence, ABCD is a parallelogram. 

Theorem- 6: The diagonals of a parallelogram bisect each other. 

To Prove: OD = OB and OC = OA

Given: ABCD is a parallelogram.

Proof: In AOD and COB

AD = CB (Opp. sides of a parallelogram) 

∠DAO = ∠BCO (Alternate Interior angle) 

∠ADO = ∠CBO (Alternate Interior angle) 

AOD  ≅  COB (ASA)

OD = OB and OC = OA (CPCT) 

Theorem-7: If the diagonals of a parallelogram bisect each other, then it is a parallelogram. 

To Prove: ABCD is a parallelogram.

Given: OD = OB and OC = OA

Proof: In AOD and COB

OD = OB  

OC = OA

∠AOD = ∠COB (Vertically opposite angle) 

AOD ≅  COB (SAS) 

∠DAO = ∠BCO (CPCT)

∠ADO = ∠CBO (CPCT)

Hence, ABCD is a parallelogram by the inverse of the alternate interior angle theorem. 

Theorem-8: Midpoint Theorem: The line segment joining the midpoints of any 2 sides of a triangle is parallel to the third side. 

To Prove: DE parallel to BC and DE = ½ BC 

Given: E and D are midpoints of AC and AB, respectively.

Construction: Draw CF parallel to AB and extend DE to point F. 

Proof: In ADE and CFE

∠ 3 = ∠ 4 (Alternate interior angle) 

∠ 1 = ∠ 2 (Vertically opposite angle) 

AE = CE (E is the midpoint of AC) 

ADE ≅   CFE (ASA)

DE = EF (CPCT) 

AD = CF (CPCT) 

AD = BD (D is the midpoint of AB) 

Hence, BD = CF 

CF is parallel to BD by construction. Hence, ABCD is a parallelogram. Meaning, 

DE is parallel to BC. 

DF = BC (Opposite sides of parallelogram) 

DE = ½ DF (as DE = EF) 

DE = ½ BC. 

Theorem-9: The line drawn through the mid-point of one side of a triangle, parallel to another side, bisects the third side. (Inverse of midpoint theorem) 

To Prove: F is the midpoint of AC. or AF=CF.

Given: E is the midpoint of AB, and EF is parallel to BC. 

Construction: Extend EF to point D. Construct CM parallel to AB.

Solution:

ED parallel to BC and CD parallel to BE. Hence, BCDE is a parallelogram. 

AE = BE (E is the midpoint of AB) 

BE = CD (Oppo. sides of a parallelogram) 

Hence, AE = CD 

∠AFE = ∠CFD (Vertically opposite angle) 

∠EAF = ∠DCF (Alternate interior angle) 

AEF ≅   CDF (AAS)

AF = CF (CPCT) 

Also Read: Understanding Quadrilaterals

4.0Key Features of CBSE Maths Notes for Class 9 Chapter 8

• These notes are aligned with the latest CBSE curriculum. 
• Visual aid is provided to get a better understanding of the chapter. 
• A step-by-step guide for solving complex problems.