THEOREMS AND PROBLEMS ON PARALLELOGRAMS – EXERCISE 4.3.2- Class IX

  1. Suppose ABCS is a parallelogram and the diagonals intersect at E. Let PEQ be a line segment with P on AB and Q on CD. Prove that PE = EQ.

Solution:

Given ABCD is a parallelogram in which AC and BD intersect at E. The line segment PEQ meet AB at P and CD at Q.

To prove: PE = EQ

THEOREMS AND PROBLEMS ON PARALLELOGRAMS - EXERCISE 4.3.2- Class IX

 

Proof 
Statementreason
In ΔAPE and ΔCQEDiagonals AC and BD bisect each other
AE = CEDiagonals AC and BD bisect each other
∟EAP = ∟CEQAP||QC, AC transversal alternate angles
∟AEP = ∟CEQvertically opposite angles
⸫ ΔAPE ≈ ΔCQEASA
⸫ PE = EQC.P.C.T

  1. Let ABCD be a parallelogram. Let BP and DQ be perpendiculars respectively from B and D on to AC. Prove that BP = CQ.

Solution:

THEOREMS AND PROBLEMS ON PARALLELOGRAMS - EXERCISE 4.3.2- Class IX

Proof 
Statementreason
In a parallelogram ABCD BP  AC  DO  ACGiven
In ΔADQ and ΔCBP

AD = CD

Opposite sides of a parallelogram
∟DAQ = ∟BCPAD || BC, AC

Transversal alternate angles

∟DQA = ∟BPC = 90°Given
⸫ ΔADQ = ΔCBPSAA
⸫BP = DQC.P.C.T

  1. Prove that in a rhombus, the diagonals are perpendicular to each other.

Solution:

Given: ABCD is a rhombus; AC and BD intersect at E.

THEOREMS AND PROBLEMS ON PARALLELOGRAMS - EXERCISE 4.3.2- Class IX

To prove: AC ⊥ BD

Proof:  
StatementReason
In rhombus, ABCD

AC and BD

Intersect each Other at E

Given
In ΔABE and ΔADEsides of a rhombus.
AB = AD

BE = DE

AE common

Diagonals bisect each other

 

⸫ ΔABE = ΔADESSS
⸫  AEB + AEDCPCT
∟AEB + ∟AED = 180˚Linear pair
⸫  AED = AED = 180/2 = 90˚ 
⸫  AC and BD are perpendicular to each other. 

  1. Suppose in a quadrilateral the diagonals bisect each other perpendicularly. Prove that the quadrilateral is a rhombus.

Solution:

THEOREMS AND PROBLEMS ON PARALLELOGRAMS - EXERCISE 4.3.2- Class IX

 

Data: ABCD is a quadrilateral, DB ⊥ AC,

AE = EC, BE = ED

To prove: ABCD is a rhombus

Proof: 
In Δ ADE and Δ DEC,

AE = EC

(data)
DE = ED(common side)
∟AED = ∟DEC90˚
⸫Δ AED = Δ DEC(RHS)
⸫ AD = DC(CPCT)
  
  

 

Similarly we can prove

AD =AB

DC = BC

So we can conclude

AB = BC = DC = AD

⸫ ABCD is a quadrilateral with equal sides and has perpendicularly bisecting diagonals.

⸫ ABCD is a Rhombus.


  1. Let ABCD be a quadrilateral in which ΔABD ≅ ΔBAC. Prove that ABCD is a parallelogram.

Solution:

5

Data: ABCD is a Quadrilateral

ΔABD ΔBAC

To prove: ABCD is a parallelogram 

BD = AC                                           (CPCT)

⸫  Area of ΔABD = Area of ΔBAC

And both the triangles stand on same base AB.

⸫  DC | | AB

But AD | | BC

⸫  AD | | BC

Hence opposite sides are parallel ABCD is a parallelogram


 

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