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.

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, $\mathrm{\Delta }$$$DAB$$ $\cong$ $\mathrm{\Delta }$$$BCD$$.

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 $$O$$.

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, $\mathrm{\Delta }$$$OAB$$$\cong$$\mathrm{\Delta }$$$OCD$$.

This implies, $$OA = OC$$ and $$OB = OD$$.

As they are equal, diagonals of a parallelogram bisect each other.