- 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.
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
|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|
- Let ABCD be a parallelogram. Let BP and DQ be perpendiculars respectively from B and D on to AC. Prove that BP = CQ.
|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.
Given: ABCD is a rhombus; AC and BD intersect at E.
To prove: AC ⊥ BD
|In rhombus, ABCD|
AC and BD
Intersect each Other at E
|In ΔABE and ΔADE||sides of a rhombus.|
|AB = AD|
BE = DE
|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.
Data: ABCD is a quadrilateral, DB ⊥ AC,
AE = EC, BE = ED
To prove: ABCD is a rhombus
|In Δ ADE and Δ DEC,|
AE = EC
|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.
- Let ABCD be a quadrilateral in which ΔABD ≅ ΔBAC. Prove that ABCD is a parallelogram.
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