- Is a parallelogram a rectangle? Can you call a rectangle a parallelogram?
Solution:
A parallelogram becomes a rectangle when its angles measure 90° each. Therefore a parallelogram need not be a rectangle. But a rectangle is a parallelogram
- Prove that bisectors of two opposite angles of parallelogram are parallel.
Given: ABCD is a | | gm AE and CF bisect DAB and DCB respectively
To prove: AE | | CF
Proof:
Statement Reason
ABCD is | |gm given
∴ DC | | AB
∠CDA + ∠DAB = 180° Co-interior angles
∴ ∠DAB = 180° – ∠CDA are supplementary
But ∠DAE = 1/2 ∠DAB ∵ AE bisects DAB
∴ ∠DAE = 1/2 [180° – ∠CDA]
∠DAE = 90° – 12 [180 – ∠CDA] ……. (1)
In Δle ADE
∠DEA = 180° – (∠DAE + ∠ADE )
= 180° – [90°– 12 ∵ ∠ ADE = ∠CDA + ∠ADE ∠CDA from equation(1)
= 180° – [90°– 1/2 ∠CDA + ∠CDA] [∵∠CDA – 1/2 ∠CDA = ∠CDA]
= 180° – [90°– 1/2 ∠CDA]
= 180° – 90°– 1/2 ∠CDA
∠DAE = 90° – 1/2 ∠CDA …….. (2)
From (1) and (2)
∠DAE = ∠DEA …….. (3)
Further, ∠ECF = 1/2 ∠DCB, But ∠DCB = ∠DAB
∠ECF = 1/2 ∠DEA
ECF = ∠DAF But from (3) DAE = DEA
∠ECF = ∠DEA
But these are corresponding angles
∴ AE | | FC
- Prove that if the diagonals of a parallelogram are perpendicular to each other, the parallelogram is a rhombus.
Solution:
Given: ABCD is a parallelogram. Diagonals AC and BD intersect at right angles.
To prove: ABCD is a rhombus
Proof: | reason |
ABCD is parallelogram | given |
AC ⊥ BD | |
In ΔAOD and ΔCOD, | |
OD is common | |
AO = CO | Diagonals bisect each other |
∠AOD = ∠COD = 90° | given |
∴ΔAOD ≅ ΔCOD | SAS |
∴AD = CD | C.P.C.T |
∴ABCD is a rhombus | adjacent side equal |
- Prove that if the diagonals of a parallelogram are equal then it is a rectangle.
Solution:
Given: ABCD is a parallelogram. Diagonals AC and BD are equal.
To prove: ABCD is a rectangle.
Proof: | reason |
In ΔDAB and ΔCBA | Opposite sides of a parallelogram. |
DA = CB | |
AB = BA | |
DB =CA | given |
∴ΔDAB ≅ ΔCBA | SSS |
∴∠DAB = ∠CBA | CPCT |
But ∠DAB + ∠CBA = 180° | AD | | BC |
∴∠DAB = ∠CBA = 180/2 = 90° | angles |
∴ABCD is a rectangle | an angle is a right angles |
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