Question

1. prove that the opposite sides and angles of a parallelogram are equal .

2. Prove that the diagonals of a parallelogram bisects each other .

3. prove that any two adjacent angles of a parallelogram are supplementary .

4. if a pair of opposite sides of a quadrilateral are equal and parallel , prove that it is a parallelogram .​

Answer 1

1.the opposite angles in a quadrilateral are equal, then it is a parallelogram. Assume that ∠A = ∠C and ∠B = ∠D . We have to prove that ABCD A B C D is a parallelogram. and thus, ABCD A B C D is a parallelogram

2.Prove: If a quadrilateral is a parallelogram, then the diagonals bisect each other.

Given: Parallelogram ABCD with diagonals BD and AC intersecting at point M.

Prove: Segment AC and BD bisect each other.

3.

Let ABCD be a parallelogram

Problems on Parallelogram

Then, AD ∥ BC and AB is a transversal.

Therefore, A + B = 180° [Since, sum of the interior angles on the same side of the transversal is 180°]

Similarly, ∠B + ∠C = 180°, ∠C + ∠D = 180° and ∠D + ∠A = 180°.

Thus, the sum of any two adjacent angles of a parallelogram is 180°.

Hence, any two adjacent angles of a parallelogram are supplementary.

4.

Answer 2

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1. prove that the opposite sides and angles of a parallelogram are equal .

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Let ABCD be a parallelogram . Join BD

In ∆ABD and ∆CDB , we have

(1)⠀⠀∠ABD = ∠CDB⠀ [ Alt . ∠s, AB || DC ]

(2)⠀⠀BD = DB⠀⠀⠀⠀ [ Common ]

(3)⠀⠀∠ADB = ∠CBD⠀ [ Alt . ∠s, AD || BC ]

∴ ⠀⠀∆ABD ≅ ∆CDB⠀ [ ASA congruency ]

∴ ⠀⠀AB = CD , AD = C⠀ [ CPCT ]

and ⠀∠A = ∠C⠀⠀⠀⠀⠀⠀ [ CPCT ]

From (1) & (3) ,

∠ABD + ∠CBD = ∠CDB + ∠ADB → ∠B = ∠D

Hence , opposite sides and angles of a parallelogram are equal .

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2. Prove that the diagonals of a parallelogram bisects each other .

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Let ABCD be a parallelogram whose diagonals AC

and BD intersects at O

In ∆AOB and ∆COD , we have

(1)⠀⠀∠OAB = ∠OCD⠀ [ Alt . ∠s, AB || DC ]

(2)⠀⠀AB = CD [ opposite sides of parallelogram ]

(3)⠀⠀∠OBA = ∠ODC⠀ [ Alt . ∠s, AB || DC ]

∴ ⠀⠀∆AOB ≅ ∆COD⠀ [ ASA congruency ]

∴ ⠀⠀OA = OC and OB = OD ⠀ [ CPCT ]

Hence , the diagonals of a parallelogram bisects each other .

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3. prove that any two adjacent angles of a parallelogram are supplementary .

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Let ABCD be a parallelogram.

Then , we have ∠A , ∠B ; ∠B , ∠C ; ∠C , ∠D and ∠D , ∠A as four pair of adjacent angles .

So, we have to prove that the sum of any two adjacent angles is 180° , i.e., ∠A + ∠B = 180° , ∠B + ∠C = 180° , ∠C + ∠D = 180° and ∠D + ∠A = 180° .

In parallelogram ABCD we have

AB || DC⠀⠀[ opposite sides of parallelogram ]

and AD is the transversal .

∴ ∠A + ∠D [sum of interior angle on the same side of transversal is 180°]

Similarly , it can be proved that

∠A + ∠B = 180° , ∠B + ∠C = 180° , ∠C + ∠D = 180°

Hence , any two adjacent angles of a parallelogram are supplementary .

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4. if a pair of opposite sides of a quadrilateral are equal and parallel , prove that it is a parallelogram .

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Let ABCD be a parallelogram in which AB = CD and AB || DC . Join BD

In ∆ABD and ∆CDB , we have

(1)⠀⠀⠀AB = CD⠀⠀⠀ [ Given ]

(2)⠀⠀∠ABD = ∠CDB⠀ [ Alt . ∠s, AB || DC ]

(3)⠀⠀BD = DB⠀⠀⠀⠀ [ Common ]

∴ ⠀⠀∆ABD ≅ ∆CDB⠀ [ SAS congruency ]

∴ ⠀⠀∠ADB = ∠CBD⠀⠀ [ CPCT ]

But ∠ADB = ∠CBD are alternate angles .

AD || BC

Thus , in quadrilateral ABCD,opposite sides are parallel .

Hence , quadrilateral ABCD is a parallelogram .