By definition, a
parallelogram
is a
quadrilateral
where both pairs of opposite sides are parallel.
If you're trying to show that a quadrilateral is a parallelogram,
using the definition isn't always the easiest approach.
There are many different ways that a
quadrilateral can be determined to
be a parallelogram:
defn of parallelogram: both pairs of opposite sides are parallel 
diagonals bisect each other 
both pairs of opposite sides are equal 
both pairs of opposite angles are equal 
one pair of opposite sides are both parallel and equal 
[beautiful math coming... please be patient]$\overline{AB}\parallel \overline{CD}$ and $\overline{AC}\parallel \overline{BD}$ 
$AE = ED$ and $CE = EB$ 
$AB = CD$ and $AC = BD$ 
$m\angle CAB = m\angle BDC$ and $m\angle ABD = m\angle DCA$ 
$\overline{AC}\parallel \overline{BD}$ and $AC = BD$ 
If you've been progressing through these
Topics in Geometry,
then you have all the tools you need to prove the previous theorem.
You could use a
chain of implications
together with a variety of prior results.
There are many routes you can take—here's one possibility:
both pairs of opposite sides parallel  $\Rightarrow$  both pairs of opposite angles are equal  Hint: You may need all these prior results: 
$\Rightarrow$  one pair of opposite sides are both parallel and equal  
[beautiful math coming... please be patient]$\Rightarrow$  both pairs of opposites sides are equal  
$\Rightarrow$  the diagonals bisect each other  
$\Rightarrow$  both pairs of opposite sides are parallel 
You should be able to negate each of the statements in the theorem above,
and phrase these negations in a variety of ways.
The skills for negating ‘and’ and ‘or’ sentences
were explored in
Logical Equivalences and Practice with Truth Tables.
They are reviewed here for your convenience:
How can a sentence ‘$\,A\text{ and }B\,$’ be false?
The only time an ‘and’ sentence is true is when both subsentences are true.
Therefore, an ‘and’ sentence is false when at least one of the subsentences is false.
Precisely, the truth table below shows that: [beautiful math coming... please be patient] $$ \text{not}(A\text{ and } B)\ \ \text{ is equivalent to }\ \ \bigl((\text{not }A)\text{ or }(\text{not }B)\bigr) $$
$A$  $B$  $A\text{ and }B$  $\text{not}(A \text{ and }B)$  $\text{not } A$  $\text{not } B$  $(\text{not }A) \text{ or } (\text{not }B)$ 
T  T  T  F  F  F  F 
T  F  F  T  F  T  T 
F  T  F  T  T  F  T 
F  F  F  T  T  T  T 
How can a sentence ‘$\,A\text{ or }B\,$’ be false?
An ‘or’ sentence is false only when both of the subsentences are false.
Precisely, the truth table below shows that: [beautiful math coming... please be patient] $$ \text{not}(A\text{ or }B)\ \ \text{ is equivalent to }\ \ \bigl((\text{not }A)\text{ and }(\text{not }B)\bigr) $$
$A$  $B$  $A \text{ or }B$  $\text{not}(A \text{ or }B)$  $\text{not }A$  $\text{not }B$  $(\text{not }A) \text{ and }(\text{not }B)$ 
T  T  T  F  F  F  F 
T  F  T  F  F  T  F 
F  T  T  F  T  F  F 
F  F  F  T  T  T  T 
Next, consider a sentence of the form: ‘$\,\text{For all }x\text{, }P\ $’
To understand a (true) sentence of this form,
think of a population where
each member has property $\,P\,$, as shown below:
What does it mean to say that the sentence
‘$\,\text{For all }x\text{, }P\ $’
is not true?
It means that at least one member of the population doesn't have property $\,P\,$, like this:
Of course, it could certainly be true that more than one member of the population doesn't have
property $\,P\,$, like this:
Precisely:
[beautiful math coming... please be patient] $\text{not}(\text{For all }x\text{, }P\,)$  $\text{ is equivalent to }$  $\text{There exists }x\text{ such that }(\text{not }P\,)$ 
Similar reasoning shows that:
[beautiful math coming... please be patient] $\text{not}(\text{There exists }x\text{ such that }P\,)$  $\text{ is equivalent to }$  $\text{For all }x\text{, }(\text{not }P\,)$ 
Here's an example of applying the negation tools:
These are all negations of: $\,Q\,$ is a parallelogram (i.e., both pairs of opposite sides are parallel)
These are all negations of: the diagonals of $\,Q\,$ bisect each other
These are all negations of: both pairs of opposite sides are equal
These are all negations of: both pairs of opposite angles are equal
Notice that ‘one pair of opposite sides are both parallel and equal’
can be rephrased as
‘at least one pair of opposite sides are both parallel and equal’ or
‘there exists a pair of opposite sides that are both parallel and equal’.
These are negations of: one pair of opposite sides are both parallel and equal
On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. 
PROBLEM TYPES:
