Featured Posts

CBSE Assignments class 09 Mathematics

  Mathematics Assignments & Worksheets  For  Class IX Chapter-wise mathematics assignment for class 09. Important and useful extra questions strictly according to the CBSE syllabus and pattern with answer key CBSE Mathematics is a very good platform for the students and is contain the assignments for the students from 9 th  to 12 th  standard.  Here students can find very useful content which is very helpful to handle final examinations effectively.  For better understanding of the topic students should revise NCERT book with all examples and then start solving the chapter-wise assignments.  These assignments cover all the topics and are strictly according to the CBSE syllabus.  With the help of these assignments students can easily achieve the examination level and  can reach at the maximum height. Class 09 Mathematics    Assignment Case Study Based Questions Class IX 

Theorems on Quadrilaterals Ch-8 Class-IX

 Theorems on Parallelograms Ch-8 Class-IX

Explanation of all theorems on Parallelograms chapter 8 class IX, Theorem 8.1, 8.2, 8.3, 8.4, 8.5, 8.6, 8.7, 8.8, 8.9, 8.10, Mid point theorem and its converse. All theorems of chapter 8 class IX.


Theorem 8.1: Prove that a diagonal of a parallelogram divides it into two congruent triangles.

Given: In Parallelogram ABCD, AC is the diagonal

To Prove:  △ACD ≌ △ABC


Proof: In △ACD and △ABC, 
∠1 = ∠2  .........   (Alternate angles
∠3 = ∠4  ..........  (Alternate interior angles
AC = AC ........   (Common Sides

⇒ By ASA ≌ rule  

△ACD ≌ △ABC


 Theorem 8.2: In a parallelogram, opposite sides are equal.

Given: ABCD is a parallelogram

To Prove : AB = CD and BC = AD

Proof: In ACD and ABC, 

1 = 2  .........   (Alternate angles

3 = 4  ..........  (Alternate interior angles

AC = AC ........   (Common Sides

By ASA  rule  

ACD ABC

AB = CD and BC = AD  …..  By CPCT


Theorem 8.3: If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram.

Given: In quadrilateral ABCD, AB = CD and BC = AD

To prove: ABCD is a parallelogram.

Proof:

In ACD and ABC, 

AD = BC  ..........  (Given

CD = AB .........   (Given

AC = AC ........   (Common Sides

By SSS  rule, 

ACD ABC

1 = 2  and 3 = 4  …..  By CPCT

Now 1 = ⇒ alternate angles are equal

⇒ AD BC  …….. (i)

Now 3 = ⇒ alternate angles are equal

⇒ AB CD …….. (ii)

From (i) and (ii) we  have

In quadrilateral ABCD both pair of opposite sides are parallel.

Hence ABCD is a Parallelogram.


 Theorem 8.4:  In a parallelogram opposite angles are equal.

Given: ABCD is a parallelogram
To prove: AD = BC and CD = AB

Proof: ABCD is a parallelogram

 AD BC

A + B = 180o ………. (i)

    ABCD is a parallelogram

  AB CD

B + C = 180o ………. (ii)

From (i) and (ii)

A + B = B + C

A = C

Similarly : B = D


Theorem 8.5: If in a quadrilateral each pair of opposite angles is equal, then it is a parallelogram.

Given: In quadrilateral ABCD , A = C and B = D

To prove: ABCD is a parallelogram

Proof: Since sum of all angles of a quadrilateral is equal to 360o

A + B + C + D = 360o

A + B +A + B = 360o

2(A + B) = 360o

A + B = 180o

⇒ Adjacent angles are supplementary

   AD BC

Similarly : AB CD

Since both pair of opposite sides are parallel

Hence ABCD is a Parallelogram


Theorem 8.6: The diagonals of a parallelogram bisects each other.

Given: ABCD is a Parallelogram.

To Prove: OA = OC and OB = OD

Proof: In AOD and BOC, 

1 = 2  .........   (Alternate angles

3 = 4  ..........  (Alternate angles

AD = BC ........   (Common Sides

By ASA  rule 

AOD ≌ BOC

OA = OC and OB = OD  …..  By CPCT


Theorem 8.7: If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.

Given: In quadrilateral ABCD, OA = OC and OB = OD

To Prove : ABCD is a parallelogram.


Proof: In AOD and BOC, 

1 = 2  .........   (Vertically Opposite angles

OA = OC  ..........  (Given

OB = OD ........   (Given

By ASA  rule  

AOD  BOC

3 = 4  and AD = BC  …..  By CPCT

3 = 4 ⇒ Alternate angles are equal

AD  BC

Now AD  BC and AD = BC 

⇒ One pair of opposite sides are equal and parallel

⇒ ABCD is a Parallelogram.


Theorem: 8.8  A quadrilateral is a parallelogram if one pair of opposite sides are equal and parallel.

Given: In quadrilateral ABCD, AD  BC and AD = BC 

To Prove: ABCD is a parallelogram

Proof: In ACD and △ABC, 

AD = BC ........ (Given

1 = 2  .........   (Vertically Opposite angles

AC = AC .... (Common Side

 By SAS  rule  

ACD  △ABC

3 = 4  …….. By CPCT

3 = 4 ⇒ Alternate angles are equal

AB  CD

Now AD  BC and AB  CD

⇒ Both pair of opposite sides are parallel

Hence ABCD is a parallelogram

Mid Point Theorem

 Theorem 8.9: If a line is drawn through the mid points of two sides of a triangle then the line is parallel to the third side and is half of it.

Given: In ABC, D and E are the mid points of sides AB and AC

To Prove: DE ॥ BC and DE =  BC


Construction: Extend DE to F such that DE = EF and join CF
Proof: In ADE and CEF
DE = EF  .........  (By Construction
AE = CE  ........  (E is the mid point of  AC
∠1 = ∠2  ........  (Vertically opposite angles

 By SAS  rule  

ADE  △CEF

∠3 = ∠4  ........  By CPCT

∠3 = ∠4 ⇒ Alternate angles are equal

⇒ AB ॥ CF   or  BD ॥ CF ......  (i)

AD = CF ....... By CPCT

But AD = BD ......  (Given

⇒ BD = CF   ...............  (ii)

From (i) and (ii) we have

BD = CF  and  BD ॥ CF

⇒ One pair of opposite sides are equal and parallel

∴ BCFD is a parallelogram

⇒ DF = BC and DF ॥ BC 




Hence prove the required theorem

Converse of Mid Point Theorem

Theorem 8.10:  

Statement:  In a triangle if a line is drawn through the mid point of one side and is parallel to the second side, then it bisects the third side.

Given: In ABC,  point D is the mid point of side AB and DE ॥ BC
To Prove: E is the mid point of AC
Construction: If possible let E is not the mid point of AC, so let us take another point F the mid point of AC.


Proof: In ABC, 
            D is the mid point of AB ........ (Given)
            F is the mid point of AC ......... (By Construction)
Therefore by mid point theorem  DF ॥ BC

But DE ॥ BC ........(Given)
This is possible only if DE and DF are coincident lines
⇒ E and F are coincide with each other.
⇒ E and F represents the same point.
Hence DF ॥ BC      DE ॥ BC
⇒ Point E is the mid - point of AC or AE = EC

Hence prove the required theorem



THANKS FOR YOUR VISIT
PLEASE COMMENT BELOW

Comments

Post a Comment


Breaking News

Popular Post on this Blog

Lesson Plan Maths Class 10 | For Mathematics Teacher

Lesson Plan Math Class X (Ch-13) | Surface Area and Volume

SUBSCRIBE FOR NEW POSTS

Followers