Parallelogram

Definition: Parallelogram is a quadrilateral where both pairs of the opposite s ides are parallel to each other.

In the figure shown, $AB\parallel DC \ and AD\parallel BC$

Note that AC is a transversal to $AB \ and\ DC$. Similarly, AC is also a transversal to $AB \ and\ DC$.

Properties of a Parallelogram (please refer to the above diagram):

1) Opposite sides are equal and parallel to each other. $AB\parallel DC \ and \ AB = DC$ $AD\parallel BC \ and \ AD = BC$

2) Opposite angles are equal $\angle A =\angle C \ and\ \angle D =\angle B$

3) Adjacent angles are supplementary (consecutive interior angles) $\angle A+\angle D=180^{\circ}$ $\angle C +\angle B = 180^{\circ}$ $\angle C +\angle D = 180^{\circ}$ $\angle A +\angle B = 180^{\circ}$

4) The diagonals (AC and BD) bisect each other $AO = OC$ $DO = OB$

5) Each diagonal bisect the parallelogram in two congruent triangles

AC bisects the parallelogram ABCD in $\Delta ACD\ and\ \Delta ACB$ such that in $\Delta ACD \cong \Delta ACB$

BD bisects the parallelogram ABCD in $\Delta ABD\ and\ \Delta BCD$ such that in $\Delta ABD \cong \Delta BCD$

Theorem: In a parallelogram, Opposite sides are equal

Opposite angles are equal

Each diagonal bisects the parallelogram

Proof:

Given: Parallelogram $ABCD$ such that $AB \parallel DC\ and\ AD \parallel BC$

To Prove: $AB = DC \ and\ AD = BC$ $\angle A =\angle C \ and\ \angle D =\angle B$ $\Delta ABD \cong \Delta BCD$ i.e. area of $\Delta ABD$ is equal to area of $\Delta BCD$

Similarly, $\Delta ACD \cong \Delta ACB$ i.e. area of $\Delta ACD$ is equal to area of $\Delta ACB$

Proof: $\angle 1 =\angle 2$ (alternate angles because $AB \parallel DC \ and\ DB$ is a transversal) $\angle 4 =\angle 3$ (alternate angles because $AD \parallel BC \ and\ DB$ is a transversal)

DB is common

Therefore, $\Delta ABD \cong \Delta BCD$    [A.S.A axiom]

i) Since $\Delta ABD \cong \Delta BCD$ $AB = DC \ and\ AD = BC$

ii) Since $\Delta ACD \cong \Delta ACB$ $\angle C = \angle A$ $\angle D = \angle B (since \angle 1 = \angle 2 \ and\ \angle 4 = \angle 3)$

Hence, $\angle A=\angle C \ and\ \angle D=\angle B$

Since $\Delta ABD \cong \Delta BCD$

Area of $\Delta ABD = \ Area \ of \Delta BCD$   (since congruent triangles are equal in area)

Similarly, if we were to join AC, we could use the same logic to prove $\Delta ACD \cong \Delta ACB$.

Rhombus Definition: A parallelogram where all four sides are equal is called a Rhombus. $AB \parallel DC \ and\ AD \parallel BC$ $AB = DC = AD = BC$

Properties of a Rhombus:

1) Opposite sides are parallel to each other. $AB \parallel DC$ $AD \parallel BC$

2) All sides are equal $AB = DC = AD = BC$

3) Diagonals AC and BD bisect each other at right angles $\angle AOB = \angle AOD = \angle BOC = \angle COD = 90^{\circ}$

4) Diagonal $BD \ bisect\ \angle A$  (i.e. $\angle ADO = \angle ODC$). Similarly, diagonal $AC \ bisects\ \angle A \ and \ \angle C.$

Similarly, $diagonal\ AC\ bisects\ \angle A \ and\ \angle C$.

Theorem: The diagonals of a Rhombus bisect each other at right angles.

Given: Rhombus $ABCD, AC \ and\ BD$ are diagonals.

To Prove: $\angle BOC=\angle COD=90^{\circ}$

Proof: $\angle ODC =\angle OBA$    (alternate angles) $\angle OCD =\angle OAB$  (alternate angles) $AB = DC$    (Side of a Rhombus)

Therefore $\Delta AOB \cong \Delta COD$   (A.S.A Axiom)

Hence $OD = OB \ and\ OA = OC$

Again in $\Delta ODC\ and\ \Delta OBC$ $DC = BC$  (Side of a Rhombus) $OD = OB$ (Proved above) And $OC$  is common

Hence $\Delta ODB \cong \Delta OBC$   (S.S.S axiom)

Therefore $\angle DOC =\angle COB$

Now $\angle DOC +\angle COB =180^{\circ}$

Or $\angle DOC =\angle COB =90^{\circ}$