# QUADRILATERALS – EXERCISE 4.2.1 – Class 9

1. 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

1. 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)

∠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

1. 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

1. 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