### Theory:

**Theorem I:**In a parallelogram, opposite sides are equal.

Given: A parallelogram \(ABCD\) with \(AC\) as its diagonal.

To prove: \(\Delta ABC\cong\Delta ADC\).

Proof: We know that 'Opposite sides of a parallelogram are parallel'. So, \(AB||DC\) and \(AD||BC\).

Since \(AB||DC\) & \(AC\) is the transversal.

\(∠BAC =∠DCA\)

*(Alternate angle ...1(yellow)).*Since \(AD||BC\) & \(AC\) is the transversal.

\(∠DAC =∠BCA\)

*(Alternate angle ...2(red)).*In \(△ABC\) & \(△ ADC\).

\(∠BAC =∠DCA\)

*(From alternate angle ...1).*\(∠DAC =∠BCA\)

*(From alternate angle ...2).*

\(AC=AC\)

*(common in both).*Thus, by the \(ASA\) criterion, the two triangles are congruent, which means that the corresponding sides must be equal.

*(ASA congruency). *

Hence, \(AB=CD\) & \(AD=BC\)

*(corresponding parts of congruent triangles).***Hence, it is proved**.

**Theorem II**: A diagonal of a parallelogram divides it into two congruent triangles.

Given: A parallelogram \(ABCD\) with \(AC\) as its diagonal.

To prove: \(ΔABC ≅ ΔADC\).

Proof: We know that 'Opposite sides of a parallelogram is parallel'. So, \(AB||DC\) and \(AD||BC\).

Since \(AB||DC\) & \(AC\) is the transversal.

\(∠BAC =∠DCA\)

*(Alternate angle ...1(yellow)).*Since \(AD||DC\) & \(AC\) is the transversal.

\(∠DAC =∠BCA\)

*(Alternate angle ...2(red)).*In \(△ABC\) & \(△ ADC\).

\(∠BAC =∠DCA\)

*(From alternate angle ...1).*\(∠DAC =∠BCA\)

*(From alternate angle ...2).*From the above figure, we can write as follows:

\(∠BAC +∠BCA\) \(=\) \(∠DAC + ∠DCA\)…(1)

Apply angle sum property to the triangle \(ABC\).

\(∠B+∠BAC+∠BCA =180°\)…(2)

Now apply angle sum property to the triangle \(ACD\).

\(∠D+∠DAC+∠DCA=180°\)…(3)

Comparing (2) and (3) with (1) we have:

\(∠B = ∠D\).

From the theorem I, \(AB=CD\) & \(AD=BC\).

Thus, by the \(SAS\) criterion, the two triangles \(ABC\) and \(ADC\) are congruent, which means that the corresponding sides must be equal.

**Hence, it is proved.**