Parallelogram is a quadrilateral whose opposite sides are parallel and congruent. Their opposite angles are equal and consecutive angles are supplementary. The diagonals bisect each other dividing the parallelogram into congruent triangles.

The theorems of parallelogram are as follows:


i) In a parallelogram opposite sides are congruent.

ii) In a parallelogram opposite angles are equal.

iii) In a parallelogram consecutive angles are supplementary.

iv) In a parallelogram, diagonals bisect each other.

v) The diagonal divides the parallelogram into two congruent triangles.

Proof: In a parallelogram opposite sides are congruent

Parallelogram Theorem Proof

Given: ABCD is a parallelogram
           AB//CD
           BC//AD
 
STATEMENT REASON
Angle ADB=Angle CBD Alternate Interior Angles
Angle ABD=Angle BDC Alternate Interior Angles
DB congruent DB Reflexive (Identity)
Triangle ADB congruent Triangle CBD ASA
AB congruent CD CPCTC
BC congruent DA CPCTC

Therefore, in a parallelogram ABCD opposite sides AB, CD and BC, DA are congruent to each other.

Proof: In a parallelogram, opposite angles are equal.

Parallelogram Theorem Proofs

Given: ABCD is a parallelogram 
           AB//CD
           BC//AD

Let us draw a line BD joining two points B and D

STATEMENT REASON
Angle ADB=Angle CBD Alternate Interior Angles
Angle ABD=Angle BDC Alternate Interior Angles
DB = DB Reflexive (Identity)
Triangle ADB $\cong$ Triangle CBD ASA
$\angle$ DAB = $\angle$ DCB CPCTC
$\angle$ ADB = $\angle$ CBD$\angle$ ADB + $\angle$ CDB Corresponding angles have equal measure
$\angle$ ADB + $\angle$ CDB =$\angle$ ABD + $\angle$ CBD Addition of equalities
$\angle$ ABD + $\angle$ CBD $\angle$ ADB + $\angle$ CDB = $\angle$ ADC $\angle$ ABD + $\angle$ CBD =Measure of Angle ABC Angle addition postulate
$\angle$ ADC= $\angle$ ABC Substitution
$\angle$ ADC = $\angle$ ABC Congruent angles have equal measures

Therefore, in a parallelogram opposite angles are equal.

Proof: In a parallelogram consecutive angles are supplementary

Given: ABCD is a parallelogram
           AB//CD
           BC//AD

Therefore, angle A and angle B are supplementary, angle B and angle C are supplementary, angle C and angle D are supplementary and angle D and angle A are supplementary angles. Reason being if two parallel lines are cut by a transversal, then the interior angles lying on the same side are supplementary angles.

Proof: In a parallelogram, diagonals bisect each other.

Given: ABCD is a parallelogram.

Parallelogram Theorems

Let diagonals AC and BD intersect at a point O.

Statement Reasons
AB//CD Definition of parallelogram
Angle BAO = Angle DCO Alternate Interior Angles
AB = CD Opposite sides of parallelogram
Angle ABO = Angle CDO Alternate Interior Angles
Triangle ABO $\cong$ Triangle CDO ASA postulate
AO = OC CPCTC
BO = OD CPCTC

Therefore, in a parallelogram, diagonals bisect each other.

Proof: The diagonal divides the parallelogram into two congruent triangles

Given: ABCD is a parallelogram
           AB//CD
           BC//AD

Parallelogram Theorem


$\left.\begin{matrix} \angle 1 = \angle 6\\ \angle 2 = \angle 5\\ \angle 3 = \angle 7\\ \angle 4 = \angle 8\\\end{matrix}\right\}$
If two parallel lines are cut by a transversal then the alternate interior angles are congruent

$\left.\begin{matrix}AC = AC\\ DB = DB\end{matrix}\right\}$ Reflexive property

$\left.\begin{matrix} Triangle\ ADC\ \cong\ Triangle\ CBA\\ Triangle\ ADB\ \cong\ Triangle\ CBD \end{matrix}\right\}$ ASA postulate of congruency.

Therefore, diagonal divides the parallelogram into two congruent triangles.

Area of parallelogram is the product of base and height.

Theorem of parallelogram area is as follows:

Parallelograms on the same base and between the same parallels are equal in area.

Proof: Parallelograms on the same base and between the same parallels are equal in area

Parallelogram Area Theorem

Let us consider two parallelograms ABCD and EFCD

In triangles ADE and BCF,

Angle DAE = Angle CBF (AD//BC and transversal AF passing making Angle DAE and CBF corresponding angles)

Angle DEA = Angle CFB (CF//DE and transversal AF passing making Angle DEA and CFB corresponding angles)

Therefore, Angle ADE = Angle BCF (Sum of angles in a triangle is 180 degrees)

Also, AD = BC (Property of parallelogram AD and BC opposite sides of parallelogram ABCD equal)

So, Triangle ADE congruent Triangle BCF (ASA postulate of congruency)

Therefore, area of triangle ADE=Area of triangle BCF 

Area of ABCD = Area of triangle ADE + Area of EDCB
                     = Area of triangle BCF + Area of EDCB
                     = Area of EFCD

So, parallelogram ABCD and parallelogram EFCD are equal in area
The parallelogram law of vector addition states that if two vector quantities are represented by two adjacent sides of a parallelogram, then the diagonal represents the resultant vector.

Parallelogram Theorem Vectors

$Vector\ R$ = $Vector V_{1}\ +\ Vector V_{2}$
i) If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram.

Converse of Parallelogram

Given: ABCD is a quadrilateral
           AB = CD
           AD = BC

Let us consider triangles ABD and BCD

Statement Reasons
AB=CD Given
AD=BC Given
BD=BD Common side to both the triangles

Therefore, triangle ABD congruent triangle BCD (SSS postulate of congruency)

So, Angle ABD = Angle CDB (By CPCTC)
      Angle ADB = Angle CBD (By CPCTC)

Therefore, AB//CD and AD//BC

Hence, Quadrilateral ABCD is a parallelogram (Proved)

ii) If in a quadrilateral each pair of opposite angles is equal, then it is a parallelogram.

Converse of Parallelogram Theorem

Given: ABCD is a quadrilateral

Angle A = Angle C ........(1)

Angle B = Angle D  .......(2)

Adding equation 1 and equation 2, we get

Angle A + Angle B=Angle C + Angle D

It is known to us that sum of interior angles of a quadrilateral is 360

Angle A + Angle B + Angle C + Angle D = 360

Angle A+ Angle B+ Angle A+ Angle B = 360

2 $\times$ (Angle A + Angle B) = 360

Angle A + Angle B = 180

Therefore, AD//BC (Consecutive interior angles)

Similarly, Angle A + Angle D = 180

AB//CD

Hence, quadrilateral ABCD is a parallelogram as opposite sides are parallel (Proved)

iii) Proof: If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.

Converse of Parallelogram Theorems

Let us consider triangles AOB and COD,

Given:
AO = OC
DO = OB

Angle AOB = Angle DOC (Vertically opposite angles)

Triangle AOB $\cong$ Triangle COD (SAS postulate of congruency)

Angle ABO = Angle ODC (CPCTC)

Angle ABD = Angle BDC (CPCTC)

As they are alternate interior angles made by the transversal BD intersecting AB and CD, we can say AB//CD.

Similarly let us consider the other two triangles AOD and BOC,

Given:
AO = OC
DO = OB

Angle AOD = Angle BOC (Vertically opposite angles)

Therefore, triangle AOD congruent Triangle BOC (SAS postulate of congruency)

Angle OAD = Angle OCD (CPCTC)
Angle ODA = Angle OBC (CPCTC)

As they are alternate interior angles made by the transversal AC intersecting AD and BC, we can say AD//BC

So, we can conclude that if the diagonals of a quadrilateral bisect each other making the opposite sides parallel to each other, then it is a parallelogram.