The parallelogram: definition and properties

Definition: a parallelogram is a quadrilateral in which both pairs of opposite sides are parallel.

• The vertices of a quadrilateral are written in order, around it: $\displaystyle A$, $\displaystyle B$, $\displaystyle C$ and $\displaystyle D$.
• In parallelogram $\displaystyle ABCD$ there are two pairs of parallel sides: $\displaystyle AB\parallel DC$ and $\displaystyle AD\parallel BC$.

Theorem: in parallelogram $\displaystyle ABCD$:

• the opposite sides are equal: $\displaystyle {AB}={DC}$ and $\displaystyle {AD}={BC}$;
• the opposite angles are equal: $\displaystyle {\angle A}={\angle C}$ and $\displaystyle {\angle B}={\angle D}$;
• two adjacent angles add up to $\displaystyle 180^{\circ}$, for example $\displaystyle {\angle A+\angle B}={180^{\circ}}$ – they are co-interior angles between the parallel lines $\displaystyle AD$ and $\displaystyle BC$;
• the diagonals bisect each other: the point where they cross is the midpoint of each of them.

Given: $\displaystyle ABCD$ is a parallelogram.
Prove: $\displaystyle {AB}={CD}$, $\displaystyle {BC}={DA}$, $\displaystyle {\angle ABC}={\angle CDA}$ and $\displaystyle {\angle DAB}={\angle BCD}$.

The proof below draws the diagonal $\displaystyle AC$, which cuts the parallelogram into two triangles, and shows that they are congruent:

Given: $\displaystyle ABCD$ is a parallelogram, and its diagonals cross at the point $\displaystyle O$.
Prove: $\displaystyle {AO}={CO}$ and $\displaystyle {BO}={DO}$.

The proof below shows that triangles $\displaystyle AOB$ and $\displaystyle COD$ are congruent, already using the property above: the opposite sides are equal.

In parallelogram $\displaystyle ABCD$, $\displaystyle {\angle A}={60^{\circ}}$, $\displaystyle {AB}={8}$ and $\displaystyle {BC}={7}$.
The diagonal $\displaystyle AC$ is $\displaystyle 13$.

• $\displaystyle {\angle B}={180^{\circ}-60^{\circ}}={120^{\circ}}$, since adjacent angles add up to $\displaystyle 180^{\circ}$;
• $\displaystyle {\angle C}={\angle A}={60^{\circ}}$ and $\displaystyle {\angle D}={\angle B}={120^{\circ}}$, since opposite angles are equal;
• $\displaystyle {CD}={AB}={8}$ and $\displaystyle {AD}={BC}={7}$, so the perimeter is $\displaystyle {2\cdot(8+7)}={30}$;
• the diagonals bisect each other at a point $\displaystyle O$, so $\displaystyle {AO}={OC}={6.5}$.

The points $\displaystyle A$, $\displaystyle B$, $\displaystyle C$ and $\displaystyle D$ are the vertices of quadrilateral $\displaystyle ABCD$.
It is a parallelogram, by the definition:

• $\displaystyle AB$ and $\displaystyle DC$ are horizontal: every point of $\displaystyle AB$ has the $\displaystyle y$-coordinate $\displaystyle 1$, and every point of $\displaystyle DC$ has $\displaystyle 4$. So $\displaystyle AB\parallel DC$.
• From $\displaystyle A$ to $\displaystyle D$ you move $\displaystyle 2$ right and $\displaystyle 3$ up, and the same from $\displaystyle B$ to $\displaystyle C$. The two segments have the same slope, $\displaystyle \frac{3}{2}$, so $\displaystyle AD\parallel BC$.

The side $\displaystyle AD$ is found by Pythagoras: $\displaystyle {AD}={\sqrt{2^2+3^2}}={\sqrt{13}}$. $\displaystyle {BC}={\sqrt{13}}$ too, as the property says.

$\displaystyle ABCD$ is a parallelogram.
Work these out.

$\displaystyle {\angle A}={65^{\circ}}$.
Find the other angles of the parallelogram.

Angle $\displaystyle A$ is twice angle $\displaystyle B$.
Find both angles.

The perimeter of the parallelogram is $\displaystyle 30$, and $\displaystyle {AB}={9}$.
Find $\displaystyle BC$.

The diagonals of parallelogram $\displaystyle ABCD$ cross at the point $\displaystyle O$.
$\displaystyle {AC}={14}$, $\displaystyle {BD}={10}$ and $\displaystyle {AB}={8}$.
What is the perimeter of triangle $\displaystyle AOB$?

The points $\displaystyle A$, $\displaystyle B$ and $\displaystyle C$ are three vertices of parallelogram $\displaystyle ABCD$.

Find the vertex $\displaystyle D$, and give the reason.

Find the length of the side $\displaystyle BC$ and the perimeter of the parallelogram.

$\displaystyle ABCD$ is a parallelogram, and $\displaystyle BD$ is a diagonal.
Perpendiculars are drawn to it from the vertices $\displaystyle A$ and $\displaystyle C$: $\displaystyle AE\perp BD$ and $\displaystyle CF\perp BD$.
Prove: $\displaystyle {AE}={CF}$.

$\displaystyle ABCD$ is a parallelogram.
The point $\displaystyle E$ is on the side $\displaystyle AB$, and the point $\displaystyle F$ on the side $\displaystyle CD$, so that $\displaystyle {AE}={CF}$.
Prove: $\displaystyle {DE}={BF}$.