### Theory:

Theorem I:
A diagonal of a parallelogram divides it into two congruent triangles. Given: $$ABCD$$ is a parallelogram, and $$AC$$ is a diagonal.

Diagonal divides the quadrilateral into two triangles $$ABC$$ and $$ADC$$.

To prove: $$\Delta ABC \cong \Delta CDA$$

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

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

Since $$AB \ || \ DC$$ and $$AC$$ is a transversal,

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

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

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

In $$\Delta ABC$$ and $$\Delta CDA$$:

$$\angle BAC = \angle DCA$$ [from (I)]

$$\angle BCA = \angle DAC$$ [from (II)]

$$AC = AC$$ [Common side]

Thus, $$\Delta ABC \cong \Delta CDA$$ [by $$ASA$$ congruence rule].

Therefore, the diagonal of a parallelogram divides it into two congruent triangles.
Theorem II:
In a parallelogram, opposite sides are equal. Given: $$ABCD$$ is a parallelogram, and $$AC$$ is a diagonal.

Diagonal divides the quadrilateral into two triangles $$ABC$$ and $$ADC$$.

To prove: $$AB = CD$$ and $$BC = AD$$

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

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

Since $$AB \ || \ DC$$ and $$AC$$ is a transversal,

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

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

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

In $$\Delta ABC$$ and $$\Delta CDA$$:

$$\angle BAC = \angle DCA$$ [from (I)]

$$\angle BCA = \angle DAC$$ [from (II)]

$$AC = AC$$ [Common side]

Thus, $$\Delta ABC \cong \Delta CDA$$ [by $$ASA$$ congruence rule].

Corresponding parts of congruent triangles are equal.

$$\Rightarrow AB = DC$$ and $$BC = AD$$.

Therefore, the opposite sides of a parallelogram are equal.
Theorem III:
If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram. Given: $$ABCD$$ is a quadrilateral, where $$AB = DC$$ and $$BC = AD$$.

Construction: Join the diagonal $$AC$$.

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

Proof: In $$\Delta ABC$$ and $$\Delta CDA$$:

$$AB = DC$$ [Given]

$$BC = AD$$ [Given]

$$AC = AC$$ [Common side]

Thus, $$\Delta ABC \cong \Delta CDA$$ [by $$SSS$$ congruence rule].

$$\implies \angle BAC = \angle DCA$$ [by CPCT] - - - - (I)

$$\implies \angle BCA = \angle DAC$$ [by CPCT] - - - - (II)

These angles are alternate interior angles.

We know that "the alternate interior angles are equal only when the lines are parallel".

So, $$AB \ || \ DC$$ and $$BC \ || \ AD$$.

Hence, $$ABCD$$ is a parallelogram.
Important!
CPCT - Corresponding Parts of Congruence Triangles