Properties of Parallelograms
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:
 The opposite sides of a parallelogram are equal (and conversely: if the opposite sides of a quadrilateral are equal, it is a parallelogram).
 The opposite angles of a parallelogram are equal (and conversely: if the opposite angles of a quadrilateral are equal, it is a parallelogram).
 The diagonals of a parallelogram bisect each other (and conversely: if the diagonals of a quadrilateral bisect each other, it is a parallelogram).
 If one pair of opposite sides of a quadrilateral is equal and parallel, then the quadrilateral is a parallelogram.
Theorem 1
Consider the following figure:
Proof:
First, we suppose that ABCD is a parallelogram. Compare \(\Delta ABC\) and \(\Delta CDA\):
\[\begin{align} & AC=AC\ \ \ \ \\&\left( \text{common sides}\right) \\\\ & \angle 1=\angle 4\ \ \ \ \ \ \\&\left( \text{alternate}\ \text{interior}\ \text{angles} \right) \\\\ & \angle 2=\angle 3\ \ \ \ \ \ \\&\left( \text{alternate}\ \text{interior}\ \text{angles} \right) \end{align}\]
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 Theorem 1
Suppose that ABCD is a quadrilateral in which \(AB=CD\) and \(AD=BC\). Compare \(\Delta ABC\) and \(\Delta CDA\) once again:
\[\begin{align} & AC=AC\ \ \ \ \\&\left( \text{common sides}\right) \\\\ & AB=CD\ \ \ \ \ \ \\&\left( \text{alternate}\ \text{interior}\ \text{angles} \right) \\\\ & AD=BC\ \ \ \ \ \ \\&\left( \text{given}\right) \end{align}\]
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.
Theorem 2
Consider the following figure:
Proof:
First, we suppose that ABCD is a parallelogram. Compare \(\Delta ABC\) and \(\Delta CDA\) once again:
\[\begin{align} & AC=AC\ \ \ \ \\&\left( \text{common sides}\right) \\\\ & \angle 1=\angle 4\ \ \ \ \ \ \\&\left( \text{alternate}\ \text{interior}\ \text{angles} \right) \\\\ & \angle 2=\angle 3\ \ \ \ \ \ \\&\left( \text{alternate}\ \text{interior}\ \text{angles} \right) \end{align}\]
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 Theorem 2
Suppose 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 \(AD\parallel BC\) (Why?)
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.
Think:
What is the relation between two lines intersected by a transversal when angles same side on the transversal are supplementary?
Theorem 3
Consider the following figure:
Proof:
First, let us suppose 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 Theorem 3
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.
Theorem 4
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:
\[\begin{align} & AB=CD\ \ \ \ \\&\left( \text{given}\right) \\\\ & \angle 1=\angle 4\ \ \ \ \ \ \\&\left( \text{alternate interior angles}\right) \\\\ & \angle 2=\angle 3\ \ \ \ \ \ \\&\left( \text{alternate interior angles}\right) \end{align}\]
Thus, the two triangles are congruent, which means that
\[\begin{align}\boxed{AE=EC\;\text{and}\;BE=ED}\end{align}\]
Therefore,the diagonals AC and BD bisect each other, and this further means that ABCD is a parallelogram (we have already proved this).
Summary
Let us summarize our discussion up to now. We have shown that the following statements are equivalent, that is, you can use them interchangeably:

A quadrilateral is a parallelogram

Opposite sides of a quadrilateral are equal

Opposite angles of a quadrilateral are equal

Diagonals of a quadrilateral bisect each other

One pair of opposite sides is equal and parallel
Solved examples on the properties of parallelograms
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}.
Example 2:
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.
Example 3:
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 \(\Delta DEG\).
\[\begin{align} & BG=GD\ \ \ \ \\&\left( \text{diagonals bisect each other}\right) \\\\ & \angle BGF=\angle DGE\ \ \ \ \ \ \\&\left( \text{vertically opposite angles}\right) \\\\ & \angle 1=\angle 2\ \ \ \ \ \ \\&\left( \text{alternate interior angles}\right)\text {(How?)} \end{align}\]
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.
Think:
What is the relation between two lines which are perpendicular to the same line?
Example 4:
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,
\[\begin{align}\boxed{\angle AEB=\angle AED = 90^\circ} \end{align}\]
Thus AC is perpendicular to BD.
Example 5:
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.
Example 6:
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.
Example 7:
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.