Properties of Parallelograms
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: 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\(\Delta CDA\):
1. AC = AC (common side)
2. \(\angle 1\) = \(\angle 4\) (alternate interior angles)
3. \(\angle 2\) = \(\angle 3\) (alternate interior angles)
Thus, by the ASA criterion, the two triangles are congruent, which means that the corresponding sides must be equal. Thus, AB = CD and AD = BC.
Now, we will prove the converse of this. Suppose that ABCD is a quadrilateral in which AB= CD and AD = BC. Compare \(\Delta ABC\) and \(\Delta CDA\) once again:
1. AC = AC (common side)
2. AB = CD (given)
3. AD = BC (given)
Thus, by the SSS criterion, the two triangles are congruent, which means that the corresponding angles are equal:
1. \(\angle 1\) = \(\angle 4\) è AB  CD
2. \(\angle 2\) = \(\angle 3\) è AD  BC
Theorem: 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:
1. AC = AC (common side)
2. \(\angle 1\) = \(\angle 4\) (alternate interior angles)
3. \(\angle 2\) = \(\angle 3\) (alternate interior angles)
Thus, the two triangles are congruent, which means that \(\angle B\) = \(\angle D\).Similarly, we can show that \(\angle A\) = \(\angle C\). This proves that opposite angles in any parallelogram are equal.
Now, we prove the converse of this. 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:
\(\angle A\) + \(\angle B\) + \(\angle C\) + \(\angle D\) = 360^{0}
è 2(\(\angle A\) + \(\angle B\)) = 360^{0}
è\(\angle A\) + \(\angle B\) = 180^{0}
This must mean that AD  BC (why?). Similarly, we can show that AB  CD, and thus, ABCD is a parallelogram.
Theorem: 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:

AB = CD (opposite sides of a parallelogram)

\(\angle 1\) = \(\angle 3\) (opposite interior angles)

\(\angle 2\) = \(\angle 4\) (opposite interior angles)
By the ASA criterion, the two triangles are congruent, which means that AE = EC and BE =ED. Thus, the two diagonals bisect each other.
Now, we will prove the converse of this. Suppose that the diagonals AC and BD bisect each other. Compare \(\Delta AEB\) and \(\Delta DEC\) once again. We have:

AE = EC (given)

BE = ED (given)

\(\angle AEB\) = \(\angle DEC\) (vertically opposite angles)
By the SAS criterion, the two triangles are congruent, which means that:

\(\angle 1\) = \(\angle 3\)è AB  CD

\(\angle 2\) = \(\angle 4\) è AD  BC
Thus, ABCD is a parallelogram.
Theorem: 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:
1. AB = CD (given)
2. \(\angle 1\) = \(\angle 3\) (opposite interior angles)
3. \(\angle 2\) = \(\angle 4\) (opposite interior angles)
Thus, the two triangles are congruent, which means that AE = EC and BE = ED. Therefore,the diagonals AC and BD bisect each other, and this further means that ABCD is a parallelogram (we have already proved this).
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
Example 1: If one angle of a parallelogram is 90^{0}, show that all its angles will be equal to 90^{0}.
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^{0}, while
\(\angle B\) = \(\angle D\) = 180^{0}  90^{0} = 90^{0}
Clearly, all the angles in this parallelogram (which is actually a rectangle) are equal to 90^{0}.
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 \(\Delta AED\):

AE = AE (common)

BE = ED (given)

\(\angle AEB\) = \(\angle AED\) = 90^{0} (given)
Thus, by the SAS criterion, the two triangles are congruent, which means that AB = AD.This further means that
AB = BC = CD = AD
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 \(\Delta DEG\).

BG = GD (diagonals bisect each other)

\(\angle BGF\) = \(\angle DGE\) (vertically opposite angles)

\(\angle 1\) = \(\angle 2\) (alternate interior angles) (how?)
By the ASA criterion, the two triangles are congruent, which means that BF = DE. Thus, B and D are equidistant from A.
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 \(\Delta ADC\):
1. AC = AC (common)
2. \(\angle 1\) = \(\angle 2\) (given)
3. \(\angle 3\) = \(\angle 4\) (given)
By the ASA criterion, the two triangles are congruent. This means that AB = AD. Now,compare \(\Delta AEB\) and \(\Delta AED\):
1. AE = AE (common)
2. AB = AD (shown above)
3. \(\angle 2\) = \(\angle 1\)
By the SAS criterion, the two triangles are congruent. Clearly,
\(\angle AEB\) = \(\angle AED\) = ½ × 180^{0} = 90^{0}
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^{0}
Now, consider \(\Delta AHB\). We have:
\(\angle 1\) + \(\angle 2\) = ½ (\(\angle A\) + \(\angle B\)) = 90^{0}
Thus, \(\angle 3\) = 90^{0}. 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,
\(\angle PAC\) = \(\angle BCA\) + \(\angle B\) = 2\(\angle BCA\)
è ½ \(\angle PAC\) = \(\angle DAC\) = \(\angle BCA\)
è BC  AD
Combining this with the fact given to us that CD  AB, 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 ⊥ BC and AD ⊥ DC.
Next, we note the following:
\(\angle ABD\) + \(\angle ADB\) = ½ (180^{0}) = 90^{0} (why?)
è \(\angle BAD\) = 90^{0}
Similarly, \(\angle BCD\) = 90^{0}. Thus, we have proved that all the four angles of the quadrilateral ABCD are right angles, which means that ABCD is a rectangle.