# Parallelogram, Rhombus, Square, Rectangle

*How to show that A, B, C, D are the vertices of a parallelogram?*

**Step 1**

Show that the diagonals AC and BD bisect each other.

*How will you show that A, B, C, D are the vertices of a Rhombus?*

**Step 1**

Show that the diagonals AC and BD bisect each other. (That is, A, B, C, D forms a parallelogram)

**Step 2Â **

Show that a pair of adjacent sides, for instance, AB and BC are equal.

*How will you show that A, B, C, D are the vertices of a Square? *

**Step 1**

Show that the diagonals AC and BD bisect each other.

**Step 2**

Show that two adjacent sides, for instance, AB and BC are equal.

**Step 3**

Show that the two diagonals AC and BD are equal.

*How will you show that A, B, C, D are the vertices of a Rectangle? *

**Step 1**

Show that the diagonals AC and BD bisect each other.

**Step 2**

Show that the two diagonals AC and BD are equal.

Â

Prove that the points A(âˆ’2, âˆ’1), B(1, 0), C(4, 3) and D(1, 2) are the vertices of a parallelogram.

Note that to show that a quadrilateral is a parallelogram, it is sufficient to show that the diagonals of the quadrilateral bisect each other.

Let M be the mid-point of AC, then coordinates of M are given by

Let N be the mid-points of BD, then coordinates of N are

Â

Thus the mid-point of AC is same as that of the mid-point of BD. In other words AC and BD bisect each other. Hence, ABCD is a parallelogram.

Â

Show that the points A(1, 3), B(2, 6), C(5, 7) and D(4, 4) are the vertices of a rhombus.

Note that to show that a quadrilateral ABCD is a rhombus, it is sufficient to show that

(a) ABCD is a parallelogram, that is AC and BD have the same mid-point and

(b) a pair of adjacent edges are equal, for instance AB = BC.

In present case, mid-point of AC is and mid-point of BD is

Thus ABCD is a parallelogram.

Also, AB^{2} = (2 â€“ 1)^{2} + (6 â€“ 3)^{2} = 1 + 9 = 10

and BC^{2} = (5 â€“ 2)^{2} + (7 â€“ 6)^{2} = 9 +1 = 10

Therefore, AB^{2} = BC^{2} or AB = BC.

Thus, ABCD is a parallelogram in which a pair of adjacent edges are equal.

Hence ABCD is a rhombus.

Â

Show that the points A(3, 2), B(5, 4), C(3, 6) and D(1, 4) are the vertices of a square.

Note that to show that a quadrilateral is a square, it is sufficient to show that

(a) ABCD is a parallelogram, that is, AC and BD bisect each other,

(b) a pair of adjacent edges are equal, for instance AB = BC and

(c) the diagonal AC and BD are equal.

In the present case, the midpoint of AC is and mid-point of BD is

Therefore, ABCD is a parallelogram.

Also, AB^{2} = (5 â€“ 3)^{2} + (4 â€“ 2)^{2} = 4 + 4 = 8

Â and BC^{2} = (3 â€“ 5)^{2} + (6 â€“ 4)^{2} = 4 + 4 = 8

Therefore, AB^{2} = BC^{2} or AB = BC.

Finally, AC^{2} = (3 â€“ 3)^{2} + (6 â€“ 2)^{2} = 16,

and BD^{2} = (1 â€“ 5)^{2} + (4 â€“ 4)^{2} = 16

That is, AC^{2} = BD^{2} or AC = BD. Hence, ABCD is a square.

Show that the points A(1, âˆ’1), B(âˆ’2, 2), C(4, 8) and D(7, 5) are the vertices of a rectangle.

To show that ABCD is a rectangle, it is sufficient to show that

(a) ABCD is a parallelogram, that is, AC and BD bisect each other, and

(b) the diagonal AC and BD are equal.

The mid-point of AC is and mid-point of BD is Therefore, ABCD is a parallelogram.

Also, AC^{2} = (4 â€“ 1)^{2} + (8 + 1)^{2} = 9 + 81 = 90

and BD^{2} = (7 + 2)^{2} + (5 â€“ 2)^{2} = 81 + 9 = 90.

Thus, AC^{2} = BD^{2} or AC = BD.

Hence, ABCD is a rectangle.