- Is a parallelogram a rectangle? Can you call a rectangle a parallelogram?
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
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.
Given: ABCD is a parallelogram. Diagonals AC and BD intersect at right angles.
To prove: ABCD is a rhombus
|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.
Given: ABCD is a parallelogram. Diagonals AC and BD are equal.
To prove: ABCD is a rectangle.
|In ΔDAB and ΔCBA||Opposite sides of a parallelogram.|
|DA = CB|
|AB = BA|
|∴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|