Parallelogram

Parallelogram A quadrilateral has both pairs of opposite sides parallel is called a parallelogram. In figure 7-5, ABCD is a parallelogram in which AB parallel to DC and AD parallel to BC . A parallelogram is denoted by the symbol ‘ ‘.

Properties of a parallelogram Property 1: A diagonal of a parallelogram divides it into two congruent triangles. Given: ABCD is a parallelogram To Prove: Triangle ABC ≅ Triangle ADC Construction: Join AC

Proof: In Triangle ABC and Triangle ACD , we have ∠BAC = ∠ACD

(Alternate interior angles)

∠ACB = ∠CAD

(Alternate interior angles)

AC=AC

(Common to both the triangles)

Therefore Triangle ABC ≅ Triangle ADC

(ASA criterion of congruency)

Hence, diagonal AC divides the parallelogram into two congruent triangles. Property 2: The opposite sides of a parallelogram are equal. Given: A parallelogram ABCD. To prove: AB=CD and DA=BC Construction: Join AC

Proof: Since ABCD is AB parallel to DC and DA parallel to BC .

a

parallelogram,

therefore,

Now, AD parallel to BC and transversal AC intersects them at A and C respectively.

Therefore ∠DAC = ∠BCA

(Alternate interior angles)

… (1)

Now, AB DC and transversal AC intersects them at A and C respectively.

Therefore ∠BAC = ∠DCA

(Alternate interior angles)

Now, in ADC and CBA, we have ∠DAC = ∠BCA

[From (1)]

AC=AC

(Common side to both the triangles)

∠BAC = ∠DCA

[From (1)]

Therefore By ASA criterion of congruence

Triangle ADC ≅ Triangle CBA Thus, AD=BC and DC=BA (Corresponding parts of congruent triangles are congruent).

… (2)

Property 3: Opposite angles of a parallelogram are equal. Given: ABCD is a parallelogram To prove: ∠A = ∠C and ∠B = ∠D

Proof: In parallelogram ABCD, AB parallel to DC and DA intersects them. therefore ∠A + ∠D = 180o

(Consecutive interior angles)

… (1)

Again, in parallelogram ABCD, AD parallel to BC and AB intersects them Therefore ∠A + ∠B = 180o From (1) and (2), we get

∠A + ∠D = ∠A + ∠B ∠D = ∠B ⇒ ∠B = ∠D

Similarly ∠A = ∠C Hence ∠A = ∠C and ∠D = ∠B

(Consecutive interior angles)

… (2)

Property 4: The diagonals of a parallelogram bisect each other. Given: A parallelogram ABCD in which the diagonals AC and AD intersect at O. To prove: OA=OC and OB=OB

Proof: Since ABCD is a parallelogram

Therefore AB parallel to DC

and

Therefore AD parallel to BC

Now, AB parallel to DC and AC is the transversal intersecting them at A and C respectively. Therefore ∠BAC = ∠DCA

(Alternate interior angles)

Thus, ∠BAO = ∠DCO

… (1)

As AB parallel to DC and BD intersects them at B and D respectively. Therefore ∠ABD = ∠CDB

(Alternate interior angles)

So, ∠ABD = ∠CDO

… (2)

Now, in Triangle AOB and Triangle COD , we have ∠BAO = ∠DCO

[From (1)]

AB=CD

(Opposite sides of a parallelogram)

∠ABO = ∠CDO

[From (2)]

therefore ASA congruence criterion

Triangle AOB ≅ Triangle COD Thus, OA=OC and OB=OB (Corresponding parts of congruent triangles are congruent) Therefore The diagonals of a parallelogram bisect each other. Property 5: The consecutive angles of a parallelogram are supplementary. Given: Parallelogram ABCD To prove: ∠A + ∠ABC = 180o Construction: Produce AB to P Proof: Since AD parallel to BC and AB is the transversal,

Therefore ∠A = ∠CBP

(Corresponding angles)

Now, ∠ABC + ∠CBP = 180o (They form a linear pair)

… (1) … 92)

From (1) and (2), we have

∠ABC + ∠A = 180o

( Since ∠CBP = ∠A )

or ∠A + ∠ABC = 180o So, consecutive angles of a parallelogram are supplementary Sufficient conditions for a quadrilateral to be a parallelogram The condition, which a quadrilateral must satisfy to become a parallelogram are know as ‘sufficient conditions’. Condition 1: A quadrilateral is a parallelogram if its opposite sides are equal.

Given: A quadrilateral ABCD in which AB=CD and BC=AD To prove: ABCD is a parallelogram Construction: Join AC Proof: In Triangle ABC and Triangle ADC , AB=CD

(Given)

BC=AD

(Given)

AC=AC

(Common to both the triangles)

Therefore Triangle ABC ≅ Triangle ADC (SSS criterion of congruency) Thus, ∠BAC = ∠DCA congruent)

(Corresponding parts of congruent triangles are

Therefore ∠BAC and ∠DCA are alternate interior angles. Hence AC DC Now, ∠BAC = ∠DCA

(cpctc)

Thus, these are alternate interior angles, Therefore BC parallel to AD Hence, quadrilateral ABCD is a parallelogram. Condition 2: A quadrilateral is a parallelogram if its opposite angles are equal. Given: A quadrilateral ABCD, in which ∠A = ∠C and ∠B = ∠D To prove: Quadrilateral ABCD is a parallelogram

Proof: In quadrilateral ABCD

∠A = ∠ C

(Given)

∠B = ∠D

(Given)

Therefore ∠A + ∠B = ∠C + ∠D

… (1)

But ∠A + ∠B + ∠C + ∠D = 360o (Sum of the angles of a quadrilateral = 360o )

… (2)

From (1) and (2), we have ∠A + ∠B + ∠C + ∠D = 360o

[ Since ∠A + ∠B = ∠C + ∠D ]

∠A + ∠B + ∠A + ∠B = 360o 2∠A + 2∠B = 360o 2(∠A + ∠B) = 360o 360o (∠A + ∠B) = 2 (∠A + ∠B) = 180o Therefore ∠C + ∠D = 180o

… (3) [ Since ∠A + ∠B = ∠C + ∠D ]… (4)

Since, transversal AB intersects lines AD and BC at A and B respectively, and (∠A + ∠B) = 180o

Therefore ∠A and ∠B form a pair of consecutive interior angles and consecutive interior angles are supplementary) So, AD parallele to BC Again, ∠A = ∠C Therefore By using this relation in (3), we have ∠C + ∠B = 180o

… (5)

Now, AB and DC are the lines and BC is the transversal, Therefore ∠B and ∠C form a pair of consecutive interior angles and consecutive interior angles are supplementary) Thus, AB parallel to DC

… (6)

From (5) an (6), we find that AD parallel to BC and AB parallel to DC Hence, quadrilateral ABCD is a parallelogram Condition 3: If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram Given: A quadrilateral ABCD, diagonals AC and BD intersect at O such that AO=OC and BO=OB. To prove: Quadrilateral ABCD is a parallelogram.

Proof: In Triangle AOD and Triangle COB, we have AO=OC

(Given)

OD=OB

(Given)

∠AOD = ∠COB

(Vertically opposite angles are equal)

Therefore By SAS criterion of congruence,

Triangle AOD ≅ Triangle COB So, ∠OAD = ∠OCB

… (1) (cpctc)

Now, lines AC intersects AD and BC at A and C respectively, such that ∠OAD = ∠OCB

… [From (10]

Therefore ∠OAD and ∠OCB form a pair of alternate interior angles (Therefore Alternate interior angles are equal) Thus, AD parallel to BC Similarly, we can prove that AB parallel to DC Hence quadrilateral ABCD is a parallelogram Condition 4: A quadrilateral is a parallelogram, if its one pair of opposite sides are equal and parallel. Given: A quadrilateral ABCD in which AB=CD and AB parallel to CD . To prove: ABCD is a parallelogram

Construction: Join AC. Proof: In Triangle ABC and Triangle CDA , we have AB=CD

(Given)

AC=AC

(Common to both triangles)

∠BAC = ∠DCA are equal)

( Since AB parallel to DC and alternate interior angles

Therefore Triangle ABC ≅ Triangle DCA Thus ∠ACB = ∠CAD

(SAS criterion of congruency)

(cpctc)

But these are alternate interior angles, Therefore AD Parallel to BC Thus we have AD parallel to BC (proved above) and AD CD (given). Hence ABCD is a parallelogram. Condition 5: A quadrilateral is a parallelogram if its opposite sides are parallel.

Proof: From the definition of a parallelogram, a quadrilateral whose opposite sides are parallel is called a parallelogram, the condition holds. ________________________________________________________________________