# 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:

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, then the quadrilateral is a parallelogram.

## Theorem 1

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:

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

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 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

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 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

In a quadrilateral ABCD, if one pair of opposite sides is equal and parallel, then it 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 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:

1. A quadrilateral is a parallelogram

2. Opposite sides of a quadrilateral are equal

3. Opposite angles of a quadrilateral are equal

4. Diagonals of a quadrilateral bisect each other

5. 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 90o, show that all its angles will be equal to 90o.

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 90o, 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 90o.

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.