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

 

Proof
Statement	reason
In ΔAPE and ΔCQE	Diagonals AC and BD bisect each other
AE = CE	Diagonals AC and BD bisect each other
∟EAP = ∟CEQ	AP||QC, AC transversal alternate angles
∟AEP = ∟CEQ	vertically opposite angles
⸫ ΔAPE ≈ ΔCQE	ASA
⸫ PE = EQ	C.P.C.T

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

Solution:

Proof
Statement	reason
In a parallelogram ABCD BP  AC  DO  AC	Given
In ΔADQ and ΔCBP AD = CD	Opposite sides of a parallelogram
∟DAQ = ∟BCP	AD || BC, AC Transversal alternate angles
∟DQA = ∟BPC = 90°	Given
⸫ ΔADQ = ΔCBP	SAA
⸫BP = DQ	C.P.C.T

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

Solution:

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

To prove: AC ⊥ BD

Proof:
Statement	Reason
In rhombus, ABCD AC and BD Intersect each Other at E	Given
In ΔABE and ΔADE	sides of a rhombus.
AB = AD BE = DE AE common	Diagonals bisect each other
⸫ ΔABE = ΔADE	SSS
⸫  AEB + AED	CPCT
∟AEB + ∟AED = 180˚	Linear pair
⸫  AED = AED = 180/2 = 90˚
⸫  AC and BD are perpendicular to each other.

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

Solution:

 

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 = ∟DEC	90˚
⸫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.

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

Solution:

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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