### 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****C**

**orresponding Parts of Congruence Triangles**