__The theorems of parallelogram are as follows:__

**i)** In a parallelogram opposite sides are congruent.

**ii)** In a parallelogram opposite angles are equal.

**iii)** In a parallelogram consecutive angles are supplementary.

**iv)** In a parallelogram, diagonals bisect each other.

**v)** The diagonal divides the parallelogram into two congruent triangles.

**Proof:** In a parallelogram opposite sides are congruent

**Given: **ABCD is a parallelogram

AB//CD

STATEMENT |
REASON |

Angle ADB=Angle CBD |
Alternate Interior Angles |

Angle ABD=Angle BDC |
Alternate Interior Angles |

DB congruent DB |
Reflexive (Identity) |

Triangle ADB congruent Triangle CBD |
ASA |

AB congruent CD |
CPCTC |

BC congruent DA |
CPCTC |

Therefore, in a parallelogram ABCD opposite sides AB, CD and BC, DA are congruent to each other.

**Proof:** In a parallelogram, opposite angles are equal.

**Given: **ABCD is a parallelogram

AB//CD

BC//AD

Let us draw a line BD joining two points B and D

STATEMENT |
REASON |

Angle ADB=Angle CBD |
Alternate Interior Angles |

Angle ABD=Angle BDC |
Alternate Interior Angles |

DB = DB |
Reflexive (Identity) |

Triangle ADB $\cong$ Triangle CBD |
ASA |

$\angle$ DAB = $\angle$ DCB |
CPCTC |

$\angle$ ADB = $\angle$ CBD$\angle$ ADB + $\angle$ CDB |
Corresponding angles have equal measure |

$\angle$ ADB + $\angle$ CDB =$\angle$ ABD + $\angle$ CBD |
Addition of equalities |

$\angle$ ABD + $\angle$ CBD $\angle$ ADB + $\angle$ CDB = $\angle$ ADC $\angle$ ABD + $\angle$ CBD =Measure of Angle ABC |
Angle addition postulate |

$\angle$ ADC= $\angle$ ABC |
Substitution |

$\angle$ ADC = $\angle$ ABC |
Congruent angles have equal measures |

Therefore, in a parallelogram opposite angles are equal.

**Proof:** In a parallelogram consecutive angles are supplementary

**Given:** ABCD is a parallelogram

AB//CD

BC//AD

Therefore, angle A and angle B are supplementary, angle B and angle C are supplementary, angle C and angle D are supplementary and angle D and angle A are supplementary angles. Reason being if two parallel lines are cut by a transversal, then the interior angles lying on the same side are supplementary angles.

**Proof:** In a parallelogram, diagonals bisect each other.

**Given:** ABCD is a parallelogram.

Let diagonals AC and BD intersect at a point O.

Statement |
Reasons |

AB//CD |
Definition of parallelogram |

Angle BAO = Angle DCO |
Alternate Interior Angles |

AB = CD |
Opposite sides of parallelogram |

Angle ABO = Angle CDO |
Alternate Interior Angles |

Triangle ABO $\cong$ Triangle CDO |
ASA postulate |

AO = OC |
CPCTC |

BO = OD |
CPCTC |

Therefore, in a parallelogram, diagonals bisect each other.

**Proof:** The diagonal divides the parallelogram into two congruent triangles

**Given:** ABCD is a parallelogram

AB//CD

BC//AD

$\left.\begin{matrix} \angle 1 = \angle 6\\ \angle 2 = \angle 5\\ \angle 3 = \angle 7\\ \angle 4 = \angle 8\\\end{matrix}\right\}$

If two parallel lines are cut by a transversal then the alternate interior angles are congruent

$\left.\begin{matrix}AC = AC\\ DB = DB\end{matrix}\right\}$ Reflexive property

$\left.\begin{matrix} Triangle\ ADC\ \cong\ Triangle\ CBA\\ Triangle\ ADB\ \cong\ Triangle\ CBD \end{matrix}\right\}$ ASA postulate of congruency.

Therefore, diagonal divides the parallelogram into two congruent triangles.