### Theory:

Theorem IV
In a parallelogram, opposite angles are equal.

Given: $$ABCD$$ is a parallelogram.

Construction: Join the diagonal $$AC$$.

To prove: $$\angle A = \angle C$$ and $$\angle B = \angle D$$

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

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

Since $$AB \ || \ CD$$ and $$AC$$ is a transversal.

$$\angle BAC = \angle DCA$$ [alternate interior angles] - - - - - (I)

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

$$\angle DAC = \angle BCA$$ [alternate interior angles] - - - - - (II)

$$\angle BAC + \angle DAC = \angle DCA + \angle BCA$$

$$\angle BAD = \angle DCB$$

That is, $$\angle A = \angle C$$ - - - - - - (III)

Now, join the diagonal $$BD$$.

Since $$AB \ || \ CD$$ and $$BD$$ is a transversal.

$$\angle ABD = \angle CDB$$ [alternate interior angles] - - - - - (IV)

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

$$\angle CBD = \angle ADB$$ [alternate interior angles] - - - - - (V)

$$\angle ABD + \angle CBD = \angle CDB + \angle ADB$$

$$\angle ABC = \angle ADC$$

That is, $$\angle B = \angle D$$ - - - - - - (VI)

From (III) and (VI), we get:

The opposite angles of a parallelogram are equal.
Theorem V
If in a quadrilateral, each pair of opposite angles is equal, then it is a parallelogram.

Given: $$ABCD$$ is a quadrilateral, where $$\angle A = \angle C$$ and $$\angle B = \angle D$$.

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

Proof: By angle sum property of a quadrilateral:

$$\angle A + \angle B + \angle C + \angle D = 360^\circ$$

$$\angle A + \angle B + \angle A + \angle B = 360^\circ$$ (Using given)

$$2 \angle A + 2 \angle B = 360^\circ$$

$$2 ( \angle A + \angle B) = 360^\circ$$

$$\angle A + \angle B = 180^\circ$$ - - - - (I)

Now, consider $$AB$$ is the transversal for the lines $$AD$$ and $$BC$$.

$$\angle A$$ and $$\angle B$$ are interior angles on the same side of transversal line $$AB$$.

"Two lines cut by a transversal are parallel if and only if interior angles on the same side of the transversal are $$180^\circ$$".

Since $$\angle A + \angle B = 180^\circ$$

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

Similarly, $$\angle A + \angle B + \angle C + \angle D = 360^\circ$$

$$\angle A + \angle D + \angle A + \angle D = 360^\circ$$ (Using given)

$$2 \angle A + 2 \angle D = 360^\circ$$

$$2 ( \angle A + \angle D) = 360^\circ$$

$$\angle A + \angle D = 180^\circ$$ - - - - (III)

Similarly, $$AD$$ is the transversal for lines $$AB$$ and $$DC$$.

$$\angle A$$ and $$\angle D$$ are interior angles on the same side of transversal line $$AD$$.

"Two lines cut by a transversal are parallel if and only if interior angles on the same side of the transversal are $$180^\circ$$".

Since $$\angle A + \angle D = 180^\circ$$

$$\Rightarrow AB \ || \ DC$$ - - - - - (IV)

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

$$AD \ || \ BC$$ and $$AB \ || \ DC$$

Both pair of opposite sides are parallel.

Therefore, $$ABCD$$ is a parallelogram.