**Theorem 1:** Parallelograms with the same base and between the same parallel lines are equal in area.

**Proof:** $\square$ABCD and $\square$BCEF are parallelograms with the same base and between the same parallel lines AE and BC.

The theorem is proved by two column proof using the figure below.

**Statement** | **Reason** |

1) ABCD and BCEF are two parallelograms with the same base BC and between the same parallels AE and BC. | 1) Given |

2) h is the altitude of the parallelogram ABCD and also of the parallelogram BCED. | 2) By the definition of altitude. |

3) Area of Parallelogram ABCD = $\frac{1}{2}$ X BC X h. | 3) By the definition of area of parallelogram. |

4) Area of Parallelogram BCEF = $\frac{1}{2}$ X BC X h. | 4) By the definition of area of parallelogram. |

5) Area of parallelogram ABCD = Area of parallelogram BCEF | 5) From 3 and 4. |

**Theorem 2:** A parallelogram is a rhombus if its diagonals are perpendicular.

**Given:** $\square$ABCD is a parallelogram where the diagonals AC and BD are at 90° to each other.

**Construction:** Join AC and BD to form diagonals. Let M be the point of intersection of diagonals.

**Proof:** In triangles AMD and DMC,

(i) $\angle$AMD = $\angle$DMC = 90°

(ii) AM = MC

(iii) DM is common.

Therefore, By SSA Postulate, $\Delta$AMD = $\Delta$DMC.

In particular, AD = CD by CPCT.

Since ABCD is a parallelogram, the opposite sides are congruent and so AB = CD, AD = BC.

Therefore, AB = BC = CD = AD.

Hence ABCD is a rhombus. The theorem is proved.

**Theorem 3:** A quadrilateral, with one pair of opposite sides parallel and equal is a parallelogram.

**Given:** ABCD is a quadrilateral, where AD || BC and AD = BC.

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

**Construction:** Draw the diagonal BD.

**Proof:** In triangles ABD and BDC,

(i) AD = BC (given)

(ii) BD is common

(iii) m$\angle$ABD = m$\angle$CDB (opposite interior angles)

By SAS postulate, $\Delta$BAD $\equiv$ $\Delta$BCD.

By CPCT, sides are equal and so AB = CD, m$\angle$ADB = m$\angle$CBD.

Thus we have, AB || CD.

Hence ABCD is a parallelogram. The theorem is proved.

**Theorem 4:** The diagonals of a parallelogram bisect each other.

**Given:** ABCD is a parallelogram. AC and BD are diagonals.

**Proof:** In $\Delta$AMB and $\Delta$CMD,

Since ABCD is a parallelogram, AB = BC

Also AB || CD, thus $\angle$ABM = $\angle$CDM

And $\angle$MAB = $\angle$DCM

So, by ASA criterion, $\Delta$AMB = $\Delta$CMD.

Thus, AM = CM, BM = DM.

Hence the diagonals of a parallelogram bisect each other.