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Theorems on Quadrilaterals Ch8 ClassIX
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Theorems on Parallelograms Ch8 ClassIX
Theorem 8.1: Prove that a diagonal of a parallelogram divides it into two congruent triangles. 
To Prove: △ACD ≌ △ABC
⇒ By ASA ≌ rule
△ACD ≌ △ABC

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.
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.

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

Given: In △ABC, D and E are the mid points of sides AB and AC
To Prove: DE ॥ BC and DE = BC
⇒ 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
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. 
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