A parallelogram is a quadrilateral with opposite sides parallel and equal. The opposite angles of a parallelogram are congruent. Any side of a parallelogram is called its base (in general, the bottom horizontal line is considered to be base). The perpendicular distance from the base to the opposite side is called the altitude or height of a parallelogram.

• A parallelogram with all equal sides and equal angles of 90° is a square.
• A parallelogram with all right angles is a rectangle.
• A parallelogram with all sides equal is a rhombus.

## Parallelogram Theorems

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.

To prove: ABCD is a parallelogram.

Construction: Draw the diagonal BD.

Proof: In triangles ABD and BDC,

(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.

### Trapezoid

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