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 = ∠2 ⇒ alternate angles are equal

⇒ AD ॥ BC  …….. (i)

Now ∠3 = ∠4 ⇒ 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 = 180^o ………. (i)

    ABCD is a parallelogram

∴  AB ॥ CD

∠B + ∠C = 180^o ………. (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 360^o

∠A + ∠B + ∠C + ∠D = 360^o

∠A + ∠B +∠A + ∠B = 360^o

2(∠A + ∠B) = 360^o

⇒ ∠A + ∠B = 180^o

⇒ 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

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