 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  APQC, AC transversal alternate angles 
∟AEP = ∟CEQ  vertically opposite angles 
⸫ ΔAPE ≈ ΔCQE  ASA 
⸫ PE = EQ  C.P.C.T 
 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 
 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. 
 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.
 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
1 thought on “THEOREMS AND PROBLEMS ON PARALLELOGRAMS – EXERCISE 4.3.2 Class IX”
Comments are closed.