Parallelograms

I. Introduction

Parallelograms are convex quadrilaterals formed by intersecting four lines, each pair of which is a set of parallel lines. A parallelogram has its opposite sides equal and opposite angles equal.

A parallelogram has the following special cases:

- Parallelogram with all sides equal gives a

Rhombus

- Parallelogram with all angles equal ($90^\circ$) gives a

Rectangle

- Parallelogram with equal sides and equal angles, gives a

Square

II. Theorem 1

$The\ opposite\ angles\ of\ a\ parallelogram\ are\ equal.$

$Given$

$ABCD$ is a parallelogram, with $\angle{BAC}$ opposite to $\angle{BCD}$ and $\angle{ABC}$ opposite to $\angle{ADC}$.

$Required\ To\ Prove$

$\angle{BAC} = \angle{BCD}$ and $\angle{ABC} = \angle{ADC}$

$Construction$

None.

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$Proof$

Line $AD$ is parallel to $BC$ and we can take the line $AB$ as transversal. From parallel lines rules, we know that interior consecutive angle sum up to $180^\circ$.

$\texttip{\therefore}{therefore} \angle{BAD} + \angle{ABC} = 180^\circ$

$\texttip{\therefore}{therefore} \angle{BAD} = 180^\circ - \angle{ABC}$

Again considering the parallel lines, $AB$ and $DC$ and considering $BC$ as the transversal, we can say that:

$\angle{ABC} + \angle{BCD} = 180^\circ$

$\angle{BCD} = 180^\circ - \angle{ABC}$

We see that both $\angle{BAD}$ and $\angle{BCD}$ have the same value, which is $180^\circ - \angle{ABC}$

$\texttip{\therefore}{therefore} \angle{BAD} = \angle{BCD}$

Similarly, $AB \parallel CD$ and $BC$ is a transversal,

$\angle{ABC} + \angle{BCD} = 180^\circ$

$\texttip{\therefore}{therefore} \angle{ABC} = 180^\circ - \angle{BCD}$

$BC \parallel AD$ and $DC$ is a transversal,

$\texttip{\therefore}{therefore} \angle{BCD} + \angle{ADC} = 180^\circ$

$therefore \angle{ADC} = 180^\circ - \angle{BCD}$

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We see that both $\angle{ADC}$ and $\angle{BCD}$ have the same value.

$\texttip{\therefore}{therefore} \angle{ABC} = \angle{ADC}$.

Now, we can see that:

$\angle{BAD} = \angle{BCD}$, and

$\angle{ABC} = \angle{ADC}$

So, we can conclude that the opposite angles of a parallelogram are equal.

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III. Theorem 2

$The\ diagonals\ of\ a\ parallelogram\ bisect\ each\ other.$

$Given$

A parallelogram $ABCD$, with $AC$ and $BD$ as diagonals, intersecting each other at point $O$.

$Required\ To\ Prove$

$OA = OC$ and $OB = OD$.

$Construction$

None.

$Proof$

Let us consider $\triangle s$ $AOB$ and $DOC$.

$AB \parallel DC$, and $AC$ is a transversal.

$\texttip{\therefore}{therefore} \angle{BAO} = \angle{DAO}$ (alternate interior angle)

$AB \parallel DC$, and $BD$ is a transversal,

$\texttip{\therefore}{therefore} \angle{ABO} = \angle{CDO}$ (alternate interior angle)

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We also know that opposite sides of a parallelogram are equal,

$\texttip{\therefore}{therefore} AB = DC$

Using the ASA rule of congruency, $\triangle{OAB} \texttip{\cong}{congruent to} \triangle{OCD}$

Therefore, the sides opposite, to the equal angles are equal.

$\texttip{\therefore}{therefore} OB = OD$ and $OA = OC$, hence $O$ is the midpoint of both $AC$ and $BD$.

Therefore the diagonals of a parallelogram bisect each other.

Parallel Lines -