Mathematics Class 10 Lab Manual | 21 Lab Activities

Mathematics Lab Manual Class X   lab activities for class 10 with complete observation Tables strictly according to the CBSE syllabus also very useful & helpful for the students and teachers.

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.

To prove: ABCD is a parallelogram.

Proof:

In ACD and ABC,

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

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

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

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

⇒ 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.

To Prove: ABCD is a parallelogram

Proof: In ACD and △ABC,

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 = $\frac{1}{2}$ BC

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

By SAS  rule

∠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

$\Rightarrow \frac{1}{2}DF = \frac{1}{2}BC\; and \; \frac{1}{2}DF \: ||\: \frac{1}{2}BC$

$\Rightarrow DE=\frac{1}{2}BC\: \: and\: \: DE \: ||\: \frac{1}{2}BC$

Hence prove the required theorem

Converse of Mid Point Theorem

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

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