### Theory:

Theorem VI
The diagonals of a parallelogram bisect each other.

Given: $$ABCD$$ is a parallelogram, where $$AC$$ and $$BD$$ are diagonals. $$O$$ is the point of intersection of diagonals.

To prove: $$OA = OC$$ and $$OB = OD$$

Proof: We know that "opposite sides of a parallelogram are parallel".

$$AD \ || \ BC$$ and $$AB \ || \ CD$$

Since $$AD \ || \ BC$$ and $$AC$$ is a transversal.

$$\angle DAC = \angle BCA$$ [alternate interior angles]

$$\Rightarrow \angle DAO = \angle BCO$$ - - - - - - (I)

Also, $$AD \ || \ BC$$ and $$BD$$ is a transversal.

$$\angle ADB = \angle CBD$$ [alternate interior angles]

$$\Rightarrow \angle ADO = \angle CBO$$ - - - - - - (II)

In $$\Delta AOD$$ and $$\Delta COB$$:

$$\angle DAO = \angle BCO$$ [from (I)]

$$\angle ADO = \angle CBO$$ [from (II)]

$$\angle AOD = \angle COB$$ [vertically opposite angles]

Therefore, $$\Delta AOD \cong \Delta COB$$ [by $$AAA$$ congruence rule].

Corresponding parts of congruence triangles are equal.

$$OA = OC$$ and $$OD = OB$$

Hence, proved.
Theorem VII
If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.

Given: $$ABCD$$ is a quadrilateral, where $$AC$$ and $$BD$$ are diagonals intersect at $$O$$.

$$OA = OC$$ and $$OB = OD$$

To prove: $$ABCD$$ is a parallelogram.

Proof: In $$\Delta AOD$$ and $$\Delta COB$$:

$$OA = OC$$ [Given]

$$\angle AOD = \angle BOC$$ [vertically opposite angles]

$$OD = OB$$ [Given]

Therefore, $$\Delta AOD \cong \Delta COB$$ [by $$SAS$$ congruence rule].

$$\Rightarrow \angle ADO = \angle CBO$$ [by CPCT]

$$\angle ADO$$ and $$\angle CBO$$ are alternate interior angles.

The alternate interior angles theorem states that, "the alternate interior angles are congruent when the transversal intersects two parallel lines".

Here, $$BD$$ is the transversal for the lines $$AD$$ and $$BC$$.

Since $$\angle ADO = \angle CBO$$

$$\Rightarrow AD \ || \ BC$$ - - - - - (I)

Similarly, in $$\Delta AOB$$ and $$\Delta COD$$:

$$OA = OC$$ [Given]

$$\angle AOB = \angle COD$$ [vertically opposite angles]

$$OB = OD$$ [Given]

Therefore, $$\Delta AOB \cong \Delta COD$$ [by $$SAS$$ congruence criterion].

$$\Rightarrow \angle BAO = \angle DCO$$ [by CPCT]

$$\angle BAO$$ and $$\angle DCO$$ are alternate interior angles.

$$AC$$ is the transversal lines for the lines $$AB$$ and $$CD$$.

Since $$\angle BAO = \angle DCO$$

$$\Rightarrow AB \ || \ CD$$ - - - - - (II)

From equations (I) and (II), we get:

$$AB \ || \ CD$$ and $$AD \ || \ BC$$.

Therefore, $$ABCD$$ is a parallelogram.