Theory:

Theorem I:
A diagonal of a parallelogram divides it into two congruent triangles.
 
A2.png
 
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.
 
A2.png 
 
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.
 
A3.png
 
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