A parallelogram is a convex polygon with 4 edges and 4 vertices.

It is a type of quadrilateral in which the opposite sides are parallel and equal.

Use the interactive tool below to learn more about parallelograms.

Drag the vertices to understand the relation between the different elements of a parallelogram.

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• A quadrilateral is a closed shape with 4 sides.
• All the internal angles of a quadrilateral add up to 360°

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The angles of a parallelogram are the 4 angles formed at the vertices.

• They all add up to 360\(^\circ\)
  (\(\angle A  +\angle B +\angle C +\angle D = 360^\circ\))
   
• Opposite angles are equal
  (\(\angle A = \angle C\: \text{and} \:\angle B = \angle D\))
   
• Consecutive angles are supplementary  
  • \(\angle A +\angle B= 180^\circ\)
  • \(\angle B +\angle C= 180^\circ\)
  • \(\angle C +\angle D= 180^\circ\)
  • \(\angle A +\angle D= 180^\circ\) 

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Diagonals are line segments which join the opposite vertices.

In parallelogram \(PQRS\), \(PR\) and \(QS\) are the diagonals.

• Diagonals bisect each other
       \(OQ =OS\)  and \(OR = OP \)
   
• Each diagonal divides the parallelogram into two congruent triangles.
       \(\Delta RSP\cong PQR\) 
      \(\Delta QPS\cong SRQ\)
   
• Parallelogram Law: The sum of the squares of the sides is equal to the sum of the squares of the diagonals.

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Theorems related to Parallelogram

The definition of a parallelogram is that it is a quadrilateral in which the opposite sides are parallel.

From this definition, we will show that:

1. The opposite sides of a parallelogram are equal (and conversely: if the opposite sides of a quadrilateral are equal, it is a parallelogram)
2. The opposite angles of a parallelogram are equal (and conversely: if the opposite angles of a quadrilateral are equal, it is a parallelogram)
3. The diagonals of a parallelogram bisect each other (and conversely: if the diagonals of a quadrilateral bisect each other, it is a parallelogram)
4. If one pair of opposite sides of a quadrilateral is equal and parallel, the quadrilateral is a parallelogram

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In a parallelogram, opposite sides are equal.

Conversely, if the opposite sides in a quadrilateral are equal, then it is a parallelogram.

Consider the following figure:

Proof:

We will assume that \(ABCD\) is a parallelogram.

Compare \(\Delta ABC\) and \(\Delta CDA\):

\[\begin{align}
& AC=CA \\
&\left( \text{common sides}\right)\\\\
 & \angle 1=\angle 4\\
&\left( \text{alternate}\ \text{interior}\ \text{angles} \right)\\\\
 & \angle 2=\angle 3 \\
&\left( \text{alternate}\ \text{interior}\ \text{angles} \right)
\end{align}\]

Thus, by the ASA criterion, the two triangles are congruent, which means that the corresponding sides must be equal.

Thus,

\[\begin{align}\boxed{AB=CD\;\text{and}\;AD=BC} \end{align}\]

Now, we will prove the converse of this.

Converse of the Theorem

If the opposite sides in a quadrilateral are equal, then it is a parallelogram.

Assume that \(ABCD\) is a quadrilateral in which \(AB = CD\)  and \(AD = BC\),

Compare \(\Delta ABC\) and \(\Delta CDA\) once again:

\[\begin{align}
& AC=AC\\
&\left( \text{common sides}\right) \\\\
& AB=CD \\
&\left( \text{since alternate interior angles are equal } \right)\\\\
 & AD=BC \\
&\left( \text{given}\right)
\end{align}\]

 

\[\begin{align}
& AC=AC\\
&\left( \text{common sides}\right) \\\\
& AB=CD \\
&\left( \text{since alternate interior}\\
 \text{angles are equal } \right)\\\\
 & AD=BC \\
&\left( \text{given}\right)
\end{align}\]

Thus, by the SSS criterion, the two triangles are congruent, which means that the corresponding angles are equal:

\[\begin{align} & \angle 1=\angle 4\Rightarrow AB\parallel CD\ \\ & \angle 2=\angle 3\Rightarrow AD\parallel BC\ \end{align}\]

Hence,

\[\begin{align}\boxed{ AB\parallel CD\;\text{and}\;AD\parallel BC}\end{align}\]

Thus, \(ABCD\) is a parallelogram.

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In a parallelogram, opposite angles are equal.

Conversely, if the opposite angles in a quadrilateral are equal, then it is a parallelogram.

Consider the following figure:

Proof:

First, we assume that \(ABCD\) is a parallelogram.

Compare \(\Delta ABC\) and \(\Delta CDA\) once again:

\[\begin{align}
& AC=AC \\
&\left( \text{common sides}\right) \\\\
& \angle 1=\angle 4 \\
&\left( \text{alternate interior angles} \right) \\\\
& \angle 2=\angle 3 \\
&\left( \text{alternate interior angles} \right)
\end{align}\]

Thus, the two triangles are congruent, which means that

\[\begin{align}\boxed{\angle B=\angle D} \end{align}\]

Similarly, we can show that

\[\begin{align}\boxed{\angle A=\angle C} \end{align}\]

This proves that opposite angles in any parallelogram are equal.

Now, we prove the converse of this.

Converse of the Theorem

If the opposite angles in a quadrilateral are equal, then it is a parallelogram.

Assume that \(\angle A\) = \(\angle C\) and \(\angle B\) = \(\angle D\).

We have to prove that \(ABCD\) is a parallelogram.

Consider the following figure:

We have:

\[\begin{align}\angle A + \angle B + \angle C + \angle D = \,360^\circ\\2(\angle A + \angle B) =\, 360^\circ\\\angle A + \angle B = \,180^\circ\end{align}\]

This must mean that \(AD\parallel BC\) 

Similarly, we can show that \(AB\parallel CD\)

Hence,

\[\begin{align}\boxed{ AD\parallel BC\;\text{and}\;AB\parallel CD}\end{align}\]

and thus, \(ABCD\) is a parallelogram.

Note that the relation between two lines intersected by a transversal, when the angles on the same side of the transversal are supplementary, are parallel to each other.

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In a parallelogram, the diagonals bisect each other.

Conversely, if the diagonals in a quadrilateral bisect each other, then it is a parallelogram.

Consider the following figure:

Proof:

First, let us assume that \(ABCD\) is a parallelogram.

Compare \(\Delta AEB\) and \(\Delta DEC\), we have:

\[\begin{align}
& AB=CD \\
&\left( \text{opposite sides of a parallelogram}\right)\\\\
& \angle 1=\angle 3\\
&\left( \text{alternate interior angles}\right) \\\\
& \angle 2=\angle 4\\
&\left( \text{alternate interior angles}\right)
\end{align}\]

By the ASA criterion, the two triangles are congruent, which means that

\[\begin{align}\boxed{AE=EC\;\text{and}\;BE=ED}\end{align}\]

Thus, the two diagonals bisect each other.

Now, we will prove the converse of this.

Converse of  the Theorem

If the diagonals in a quadrilateral bisect each other, then it is a parallelogram

Suppose that the diagonals AC and BD bisect each other.

Compare \(\Delta AEB\) and \(\Delta DEC\) once again. We have:

\[\begin{align} & AE=EC \\
&\left( \text{given}\right) \\\\
& BE=ED \\
&\left( \text{given}\right) \\\\
& \angle AEB=\angle DEC\\
&\left( \text{vertically opposite angles}\right)
 \end{align}\]

By the SAS criterion, the two triangles are congruent, which means that:

\(\angle 1\) = \(\angle 3\)

\(\angle 2\) = \(\angle 4\)

Hence,

\[\begin{align}\boxed{AB\parallel CD\;{\rm{and}}\;AD\parallel BC} \end{align}\]

Thus, \(ABCD\) is a parallelogram.

---

Consider the following figure:

It is given that \(AB=CD\) \(\)and \(AB || CD \)

We have to prove that \(ABCD\) is a parallelogram.

Proof:

Compare \(\Delta AEB\) and \(\Delta DEC\). We have:

\[\begin{align}
& AB=CD\\
&\left( \text{given}\right)\\\\
& \angle 1=\angle 3 \\
&\left( \text{alternate interior angles}\right)\\\\
& \angle 2=\angle 4\\
&\left( \text{alternate interior angles}\right)
 \end{align}\]

Thus, the two triangles are congruent, which means that

\[\begin{align}\boxed{AE=EC\;\text{and}\;BE=ED}\end{align}\]

Therefore,the diagonals AC and BD bisect each other, and this further means that \(ABCD\) is a parallelogram.

Solved Examples

Example 1

 

 

If one angle of a parallelogram is 90^o, show that all its angles will be equal to 90^o.

Solution:

In any parallelogram, the alternate angles are equal and the adjacent angles are supplementary.

Consider the parallelogram \(ABCD\) in the following figure, in which \(\angle A\) is a right angle:

Thus, \(\angle C\) must also be 90^o, while

\(\begin{align}\angle B = \angle D &=\,180^\circ - \;90^\circ \\\\&=\,90^\circ\end{align}\)

Hence,

\[\begin{align}\boxed{\angle A=\angle B=\angle C=\angle D = 90^\circ} \end{align}\]

Clearly, all the angles in this parallelogram (which is actually a rectangle) are equal to 90^o.

\(\therefore\) when one angle is 90o, the parallelogram is a rectangle

Example 2

 

 

If the diagonals of a parallelogram are equal, then show that it is a rectangle.

Solution:

Consider a parallelogram \(ABCD\) with digonals \(AC\) and \(BD\).

Given: 

• \(ABCD\) is a parallelogram
• \(AC\) = \(BD\)

To prove: \(ABCD\) is a rectangle.

In \(\Delta ABC\)  and  \( \Delta ABD\),

• \(AB = AB\) (common side)
• \(AC = BD\) (given)
• \(BC = AD\) (opposite sides of a parallelogram are equal) 

Therefore,

\( \Delta ABC  \cong  \Delta BAD \)

[ by SSS congruence axiom]

By CPCT, 

\(\angle ABC = \angle BAD\) -------- (1)

 But also, 

\( \angle ABC + \angle BAD\) = 180°

(co - interior angles are supplementary)

\(\implies\) \( \angle ABC\) + \( \angle ABC \) = 180° (from (1))

Therefore,

\(\angle ABC\) = 90°

\(\implies\) All angles are 90° (corresponding angles are equal and consecutive angles are supplementary in a parallelogram)

\(\therefore\) ABCD is a rectangle

Example 3

 

 

What is the difference between the opposite angles of a parallelogram?

a) 180°
b) 90°
c) 0°
d) 360°

Solution:

In a parallelogram, the opposite sides and opposite angles are equal.

Therefore, the difference between the opposite angles of a parallelogram is:

\(x-x = 0^\circ \)

\(y-y = 0^\circ \)

\(\therefore\) Difference between opposite angles of a parallelogram is 0°

Example 4

 

 

In a quadrilateral \(ABCD\), the diagonals \(AC\) and \(BD\) bisect each other at right angles.

Show that the quadrilateral is a rhombus.

Solution:

Consider the following figure:

First of all, we note that since the diagonals bisect each other, we can conclude that \(ABCD\) is a parallelogram.

Now, let us compare \(\Delta AEB\) and \(\Delta AED\):

\[\begin{align} & AE=AE\ \ \ \ \\&\left( \text{common}\right) \\\\ & BE=ED\ \ \ \ \ \ \\&\left( \text{given}\right) \\\\ & \angle AEB=\angle AED=\,90^\circ\ \ \ \ \ \ \\&\left( \text{given}\right) \end{align}\]

Thus, by the SAS criterion, the two triangles are congruent, which means that

\(AB\) = \(CD\)

This further means that

\[\begin{align}\boxed{ AB=BC=CD=AD} \end{align}\]

Clearly, \(ABCD\) is a rhombus.

\(\therefore\) Parallelogram ABCD is a rhombus

Example 5

 

 

\(ABCD\) is a quadrilateral in which the diagonals bisect each other.

Show that \(B\) and \(D\) are equidistant from \(AC\).

Solution:

Since its diagonals bisect each other, \(ABCD\) is a parallelogram.

Consider the following figure:

Essentially, we have to show that

\(BF\) = \(DE\)

Compare \(\Delta BFG\) with \(\Delta DEG\).

\[\begin{align} & BG=GD\ \ \ \ \\&\left( \text{diagonals bisect each other}\right) \\\\ & \angle BGF=\angle DGE\ \ \ \ \ \ \\&\left( \text{vertically opposite angles}\right) \\\\ & \angle 1=\angle 2\ \ \ \ \ \ \\&\left( \text{alternate interior angles}\right) \end{align}\]

By the ASA criterion, the two triangles are congruent, which means that:

\[\begin{align}\boxed{ BF=DE} \end{align}\]

Thus, \(B\) and \(D\) are equidistant from \(A\).

 Note: Two lines which are perpendicular to the same line are parallel to each other.

\(\therefore\) B and D are equidistant from AC

Example 6

 

 

If in a quadrilateral \(ABCD\), \(AC\) bisects angles A and C, show that \(AC\) is perpendicular to \(BD\)

Solution:

Consider the following figure:

Compare \(\Delta ABC\) and \(\Delta ADC\):

\[\begin{align} & AC=AC\ \ \ \ \\&\left( \text{common}\right) \\\\ & \angle 1=\angle 2\ \ \ \ \ \ \\&\left( \text{given}\right) \\\\ & \angle 3=\angle 4\ \ \ \ \ \ \\&\left( \text{given}\right) \end{align}\]

By the ASA criterion, the two triangles are congruent.

This means that:

\(AB\) = \(AD\)

Now, compare \(\Delta AEB\) and \(\Delta AED\):

\[\begin{align} & AE=AE\ \ \ \ \\&\left( \text{common}\right) \\\\ & AB=AD\ \ \ \ \ \ \\&\left( \text{shown above}\right) \\\\ & \angle 2=\angle 1\ \ \ \ \ \ \\&\left( \text{given}\right) \end{align}\]

By the SAS criterion, the two triangles are congruent. Clearly,

\[\begin{align}\angle AEB = &\angle AED\\\\= &\frac{1}{2} \times 180^\circ \\ \\= &\,90^\circ\end{align}\]

Hence,

\[\begin{align}\boxed{\angle AEB=\angle AED = 90^\circ} \end{align}\]

\(\therefore\) AC is perpendicular to BD

Example 7

 

 

Prove that the bisectors of the angles in a parallelogram form a rectangle.

Solution:

Consider the following figure, in which \(ABCD\) is a parallelogram, and the dotted lines represent the (four) angle bisectors.

We have to show that \(EFGH\) is a rectangle:

We can show this by proving that each of the four angles of \(EFGH\) is a right angle.

In parallelogram \(ABCD\), we have:

\(\angle A + \angle B = 180^\circ \)

Now, consider \(\Delta AHB\).

We have:

\(\begin{align}\angle 1 + \angle 2 =& \frac{1}{2}\left( {\angle A + \angle B} \right)\\\\ =&\,\ 90^\circ\end{align}\)

Thus,

\[\begin{align}\boxed{\angle 3 = 90^\circ} \end{align}\]

Similarly, we can prove that each of the other three angles of quadrilateral \(EFGH\) is a right angle.

This means that \(EFGH\) is a rectangle.

\(\therefore\) Bisectors of the angles in a parallelogram form a rectangle

Example 8

 

 

ABC is an isosceles triangle in which AB = AC.

AD bisects exterior angle PAC and CD is parallel to AB, as shown below:

Show that:

1. \(\angle DAC\) = \(\angle BCA\)

2. ABCD is a parallelogram

Solution:

We note that since \(\Delta ABC\) is isosceles (with \(AB = AC\)), \(\angle B\) = \(\angle BCA\).

Now, by the exterior angle theorem,

\(\begin{align}\angle PAC &= \angle BCA + \angle B \\\\&=2\angle BCA\end{align}\)

\(\begin{align}\frac{1}{2}\angle PAC &= \angle DAC \\ \\&= \angle BCA\end{align}\)

Therefore,

\[\begin{align}\boxed{ \angle DAC = \angle BCA} \end{align}\]

Hence,

\(BC\) = \(AD\)

Combining this with the fact given to us that

\(CD\) = \(AB\)

Hence,

\[\begin{align}\boxed{BC\parallel AD\;{\rm{and}}\;CD\parallel AB} \end{align}\]

Thus, we conclude that ABCD is a parallelogram.

\(\therefore\) ABCD is a parallelogram.

Example 9

 

 

Two parallel lines are intersected by a transversal.

Consider the quadrilateral ABCD formed by the bisectors of the interior angles, as shown below:

Show that ABCD is a rectangle.

Solution:

We first note that the two bisectors of any pair of lines are perpendicular to each other (if you are unsure about this, take a moment to prove it). Thus,

\(AB \bot BC{\rm{\;\;and\;\; }}AD \bot DC\)

Next, we note the following:

\(\begin{align}\angle ABD + \angle ADB &= \frac{1}{2} \times 180^\circ\,\,\text{(Why?)}\\ \\&=\, 90^\circ\end{align}\)

\[\begin{align}\boxed{\angle BAD = 90^\circ} \end{align}\]

Similarly,

\[\begin{align}\boxed{\angle BCD = 90^\circ} \end{align}\]

Thus, we have proved that all the four angles of the quadrilateral ABCD are right angles, which means that ABCD is a rectangle.

\(\therefore\) ABCD is a rectangle

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