Theory:

There are some properties which are always true for any parallelogram. 
Let's see those properties with explanation.
In a parallelogram, the following properties are true:
  1. the opposite sides are equal in length.
  2. the opposite angles are equal in measure.
  3. the adjacent angles are supplementary.
  4. the diagonals bisect each other.
Proof:
 
Let's prove (1) and (2).
 
Let \(ABCD\) be a parallelogram. 
 
Theory_2_1_1.png
 
Draw a diagonal \(BD\) and denote the interior angles as \(∠1, ∠2, ∠3\) and \(∠4\).
 
Since \(ABCD\) is a parallelogram, \(AD\) is parallel to \(BC\) and \(AB\) is parallel to \(CD\).
 
Consider the parallel lines \(AD\) and \(BC\) and take the diagonal \(BD\) as transversal.
 
Here \(∠2 = ∠4\) by alternate interior angle property [Alternate interior angles are equal in measure].
 
Now consider parallel lines \(AB\) and \(CD\) and take the diagonal \(BD\) as transversal.
 
Here \(∠1 = ∠3\) by alternate interior angle property [Alternate interior angles are equal in measure].
 
So we have  \(∠1 + ∠2 = ∠3+∠4\). That is, \(∠B = ∠D\).
 
Consider the triangles \(DAB\) and \(BCD\) with the common side \((BD = BD)\).
 
Now we have \(∠2 = ∠4\), \(∠1 = ∠3\) and the common side \((BD = BD)\).
 
By \(ASA\) criterion of congruence, Δ\(DAB\)  Δ\(BCD\).
 
That is, they are congruent triangles.
 
Therefore, \(AD = BC, AC = CD\) and \(∠A= ∠C\).
 
Hence, in parallelogram \(ABCD\), we have \(AD = BC, AC = CD, ∠A = ∠C\) and \(∠B = ∠D\).
 
This proves (1) and (2).
 
From (2), it is obvious that \(∠A = ∠C\) and \(∠B = ∠D\).
 
Let's prove (3).
 
Let the measure of angles are\(∠A = ∠C = x\) and \(∠B = ∠D = y\).
 
We have the property that sum of all interior angles of a quadrilateral is \(360°\).
 
That is, \(∠A + ∠B +∠C + ∠D = 360°\).
 
Substituting the taken values in the above equation.
 
\(x + y + x + y = 360°\)
 
\(2x +2y = 360°\)
 
\(2(x + y) = 360°\)
 
\(x + y  = 180°\).
 
That is, we can write this as \(∠A + ∠B = 180°\), \(∠B + ∠C = 180°\), \(∠C + ∠D = 180°\) and \(∠A + ∠D = 180°\).
 
It proves the property (3).
 
Let's prove (4).
 
Consider a parallelogram \(ABCD\). Draw its diagonals \(AC\) and \(BD\). Let the intersection point of the diagonals be \(O\).
 
10.PNG
 
In triangle \(AOB\) and \(COD\), we have:
 
\(AB = CD\) as opposite sides are equal in parallelogram.
 
\(∠AOB = ∠COD\) [Because vertically opposite angles are equal].
 
Here \(AB\) is parallel to \(CD\), so \(∠OAB = ∠DCO\).
 
By AAS criterion of congruence, Δ\(OAB\)Δ\(OCD\).
 
This implies, \(OA = OC\) and \(OB = OD\).
 
As they are equal, diagonals of a parallelogram bisect each other.