# What are the types of quadrilaterals?

Quadrilaterals are part of a plane enclosed by four sides (quad means four and lateral means side). All quadrilaterals have exactly four sides and four angles, and they can be sorted into specific groups based on lengths of their sides or measures of their angles. What they have in common is that in every quadrilateral the sum of the measures of all interior angles is equal to $360^{\circ}$.

Their vertices are marked with capital letters and sides with small letters.

Angles in vertices $A, B, C$ and $D$ are usually marked in order with: $\alpha, \beta, \gamma, \delta$ (alpha, beta, gamma, delta).

Remember that in triangles, the sum of  the measures of all exterior angles is equal to $360^{\circ}$ (remember, exterior angle is a supplementary angle to a certain interior angle).

This will also be true for quadrilaterals. The sum of all measures of exterior angles in quadrilaterals is always equal to $360^{\circ}$.

Diagonals are lines that connect opposite angles.

The division of quadrilaterals according to perpendicularity diagonals and parallel sides:

First group of quadrilaterals is a scalene quadrilateral. Scalene quadrilateral is a quadrilateral that doesn’t have any special properties; the sides  and angles have different lengths and measures.

Quadrilaterals which have one pair of parallel sides are called trapezoids. Sides that are parallel are called bases of a trapezoid, and ones that are not parallel are called legs.

Trapezoids whose legs are of equal length are called isosceles trapezoids.

Diagonals of a isosceles trapezoids are congruent.

Height or altitude of a trapezoid is the length of a line that is perpendicular to a base and runs through opposite vertex. Altitude of a trapezoid will be equal no matter from which vertex we draw it. If we are drawing an altitude from larger base, we simply extend the shorter base.

Theorem.

If $\alpha$ is an angle in vertex $A$, $\beta$ in vertex $B$, $\gamma$ in vertex $C$ and $\delta$ in vertex $D$ in a trapeziod $ABCD$, then is valid:

$\alpha + \delta = 180^{\circ}$
$\beta + \gamma = 180^{\circ}$.

In other words, the angles on the same side of a leg of a trapezoid are supplementary.

Proof.

Expand the segment $\overline{AD}$ over the vertices $A$ and $D$. On the line $AD$ denote point $E$.  Since the line $AD$ is a transverse of the parallel lines $AB$ and $CD$,  then $\angle{CDE} =\alpha$ is valid . The angles $\angle{CDE}$ and $\delta$ are supplementary angles, which means that  $\angle{CDE} +\delta = 180^{\circ} \Rightarrow \alpha+ \delta = 180^{\circ}$.

Analogously, we obtained  $\beta+ \gamma = 180^{\circ}$.

A parallelogram is a quadrilateral whose opposite sides are congruent and parallel.

Altitude or a height of a parallelogram, in the label $h$, is the line segment that connects a vertex with opposite side, and is perpendicular to that side.

Theorem.

Let $ABCD$ be a parallelogram. The opposite angles in a quadrilateral $ABCD$ are congruent, and the adjacent angles  are supplementary.

Proof.

By definition,  if  $ABCD$  is a trapezoid with legs $\overline{AD}$ and $\overline{BC}$, then:

$$\beta + \gamma = 180^{\circ} \quad and \quad \alpha + \delta = 180^{\circ}.$$

If  $ABCD$  is a trapezoid with legs $\overline{AB}$ and $\overline{CD}$, then:

$$\alpha + \beta = 180^{\circ} \quad and \quad \beta + \delta = 180^{\circ}.$$

It follows $\beta = \delta$ and $\alpha = \gamma$.

Theorem.

The following statements are equivalent to each other:

1) A quadrilateral $ABCD$ is a parallelogram

2) There exists two opposite sides of a quadrilateral $ABCD$ which are congruent and parallel

3) Each two opposites sides of a quadrilateral $ABCD$ are congruent

4) Diagonals of a quadrilateral $ABCD$ bisect each other

5) Both pairs of opposite angles of a quadrilateral $ABCD$ are congruent

Each of the above statements can be an alternative definition of a parallelogram. The remaining statements we need to prove.

Proof.

$1) \Rightarrow 2)$

Let $ABCD$ be a parallelogram. Then $\overline{AB} \| \overline{CD}$ and $\overline{AD} \| \overline{BC}$.

Since line $AC$ is a traverse of parallel lines $AB$ and $CD$ ,then $\angle{ACD} =\angle{CAB}$. A line $AC$ is also a traverse of parallel lines $BC$ and $AD$ that’s $\angle{ACB}=\angle{DAC}$.

$\overline{AC}$ is also the common side of triangles $ABC$ and $CDA$. By A-S-A theorem of congruence of triangles, triangles $ABC$ and $CDA$ are congruent. It follows that $|AB| = |CD]$ and $|AD| = |BC|$.

$2) \Rightarrow 3)$

In quadrilateral  $ABCD$ let be $AB \| CD$ and $|AB| = |CD|$.

Since $AC$ is a traverse of the parallel lines $AB$ and $CD$, that is $\angle{ACD} =\angle{CAB}$. The side $\overline{AC}$ is common side of triangles $ABC$ and $CDB$. By S-A-S theorem of congruence of triangles, triangles $ABC$ and $CDB$ are congruent. It follows that $|BC| = |CD|.$

$3) \Rightarrow 4)$

In quadrilateral $ABCD$ let be $|AB|=|CD|$ and $|BC| = |CD|$, and let the point $S$ be the intersection of diagonals $\overline{AC}$ and $\overline{BD}$.

First, consider triangles $ABC$ and $CDB$. By S-S-S theorem of congruence of triangles, triangles $ABC$ and $CDB$ are congruent. It follows that $\angle{ACB} = \angle {CAD}$.

Angles $ASD$ and $BSC$ are vertical angles. If now consider triangles $ASD$ and $BSD$, it follows that $\angle{ADS} = \angle{CBS}$. Since $|BC|= |AD|$ that triangles $ASD$ and $BSC$ are congruent by A-S-A theorem of congruence triangles. It follows that $|AS| = |SC|$ and $|BS| = |SD|$ which means that point $S$ is the midpoint of $\overline{AC}$ and $\overline{BD}$.

$4) \Rightarrow 5)$

In quadrilateral $ABCD$ let the point $S$ be the midpoint of diagonals $\overline{AC}$ and $\overline{AD}$: $|AS|=|CS|$ and $|BS|=|DS|$.

Consider triangles $BCS$ and $ADS$. By S-A-S theorem they are  congruent ($|CS| = |AS|$ – $\angle{BSC}=\angle{ASB}$ (vertical angles) – $|BS|=|DS|$). It follows that $\angle{BCS}=\angle{DAS}$ and $\angle{CBS}=\angle{ADS}$.

Triangles $ABS$ and $CDS$ are also congruent by S-A-S theorem ($|AS|=|CS|$ – $\angle{ASB}=\angle{CSD}$ (vertical angles)  – $|BS|=|DS|$). It follows that $\angle{BAS}=\angle{DCS}$ and $\angle{ABS}=\angle{CDS}$.

It follows:

$$\angle{DAB} = \angle{DAS} + \angle{BAS}= \angle{BCS} + \angle{DCS} = \angle{BCD}$$

and

$$\angle{ABC} = \angle{ABS} + \angle{CBS}= \angle{CDS} + \angle{ADS} = \angle{ADC}$$.

$5) \Rightarrow 1)$

In quadrilateral $ABCD$ let be $\alpha=\gamma$ and $\beta = \delta$. That means that $\alpha+\beta = 180^{\circ}$ and $\gamma+\delta = 180^{\circ}$.

Assume that lines $AB$ and $AC$ are not parallel and let the point of intersection of these two lines be the point $E$ which is on the same side of line $BC$ and points to $A$ and $D$. Then angles $\gamma$ and $\delta$ are interior angles of triangle $CDE$, but the sum of measures of angles $\gamma$ and $\delta$ is equal to $180^{\circ}$, which is a contradiction.

If point $E$ is on the opposite side of line $BC$ and points to $A$ and $D$, then $\alpha + \beta = 180^{\circ}$, which is also a contradiction.

It follows that $AB \| AC$.

Similarly we prove that $BC \| AD$.

rhombus is a parallelogram which has at least one pair of adjacent sides of equal length.

Opposite angles are of equal measure: $\alpha = \gamma$, $\beta = \delta$ and that adjacent angles are supplementary.

Diagonals in rhombus are congruent and perpendicular.

A kite is a quadrilateral which characterizes two pairs of  sides of equal lengths that are adjacent to each other. Diagonals of a kite are perpendicular and at least one diagonal is a line of symmetry. A kite is also a tangential quadrilateral.

A rectangle is a parallelogram which at least one interior angle is right.

Diagonals in rectangles are congruent.

Square is a rectangle whose all sides are equal.

Diagonals in a square are congruent and perpendicular.

## Perimeters and areas of  quadrilaterals

Perimeter of any geometric shape is the length of is outline.

Area of any geometric shape is the surface it occupies. Unit of measure for area is $m^2$ (square meter).

$1$ square meter is equal to the surface enclosed by a square with sides $1 m$.

There are also some derived units of measure for areas, for smaller or larger shapes.

$1 km^2 = 1 km \cdot 1 km = 1000 m \cdot 1000 m = 1 000 000 m^2$

$1 m^2 = 1 m \cdot 1m$

$1 dm^2 = \frac{1}{10} m^2 \cdot \frac{1}{10} m^2 = \frac{1}{100} m^2$

$1 cm^2 = \frac{1}{100} m^2 \cdot \frac{1}{100} m^2 = \frac{1}{10 000} m^2$

$1 mm^2 = \frac{1}{1000} m^2 \cdot \frac{1}{1000} m^2 = \frac{1}{1 000 000} m^2$

Area of a square is equal to a square of length of its side.

Area of a rectangular is equal to the product of lengths of adjacent sides.

Area of a rhombus is equal to the product of length of its side and altitude. This is true because, from the picture: if we translate altitude $\overline{BE}$ into point $A$, and extend side $ED$ over vertex $D$, we will get triangle $E’DA$ which is congruent with triangle $ECB$. If we ‘translate’ triangle $ECB$ onto triangle $E’DA$ we will get a rectangular with one side $a$ and other $h$.

The same that goes for a rhombus works on a parallelogram, the area of a parallelogram is a product of its one side and altitude on that side.

Area of a trapezoid is equal to one half of a product of sum of its bases and altitude.

This formula is a result of dividing a trapezoid into a two triangles $AED$ and $BCF$, and a rectangle $EFCD$.

Now, we can write our area as the sum of smaller areas: $A_{(ABCD)} = A_{(AED)} + A_{(FBC)} + A_{(EFCD)}$.

We know that $A_{(EFCD)} =h \cdot c$.
Now we need to find $A_{(AED)}$ and $A_{(FBC)}$. If we translate side $b$ next to $AED$ we get a triangle $AHD$.

The altitude of a triangle $AHD$ is equal to the altitude of a trapezoid $ABCD$.

And the side on which this altitude is set is equal to $a – c$. This leads to a conclusion that:

$\ A_ {(AHD)} = h \cdot \frac{a – c}{2}$.

This means that:

$$A_ {(ABCD)} = A_ {(AHD)} + A_ {(HBCD)} =$$

$$=\frac{h \cdot (a – c)}{2} + c\cdot h= \frac{a \cdot h – h\cdot c + 2h \cdot c}{2} = \frac{h \cdot a + h \cdot c}{2} =$$

$$=\frac{h \cdot (a + c)}{2}$$.

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