# 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

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:

 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

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.

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.

1. 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

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:

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