- 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 |
- 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
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