Quadrilaterals

Quadrilaterals

A quadrilateral is a polygon that has four sides. The line segment sides lie within a single plane.

Quadrilaterals with non-coplanar sides are called “SKEW”.

PARALLELOGRAM:

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

THEOREMS:

-A diagonal of a parallelogram separates it into two congruent triangles.

COROLLARIES:

-The opposite angles of a parallelogram are congruent

-The opposite sides of a parallelogram are congruent

-The diagonals of a parallelogram bisect each other.

-Two consecutive angles of a parallelogram are supplementary.

Theorems:

-In a parallelogram with unequal pairs of consecutive angles, the longer diagonal lies opposite the obtuse angle.

-If two sides of a quadrilateral are both congruent and parallel, then the quadrlateral is a parallelogram.

-If both pairs of opposite sides of a quadrilateral are congruent, then the quadrilateral is a parallelogram.

-If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram.

KITE:

The $KITE$ is a quadrilateral with two distinct pairs of congruent adjacent sides.

Theorem:

In a kite, one pair of opposite angles are congruent.

RECTANGLE, SQUARE AND RHOMBUS

A rectangle is a parallelogram that has a right angle.

COROLLARIES:

-All angles of a rectangle are right angles.

-The diagonals of a rectangle are congruent.

A square is a rectangle that has two congruent adjacent sides.

COROLLARIES

-All sides of a square are congruent

A rhombus is a parallelogram with two adjacent congruent sides.

COROLLARY:

All sides of a rhombus are congruent.

THEOREM:

The diagonals of a rhombus are perpendicular.

THE TRAPEZOID:

A trapezoid is a quadrilateral with exactly two parallel sides.

Theorems:

-The base angles of an isosceles trapezoid are congruent.

-The length of the median of a trapezoid equals one-half the sum of the lengths of the two bases.

$median=\frac{b_{1}+b_{2}}{2}$

-The median of a trapezoid is parallel to each base

-If two base angles of a trapezoid are congruent, the trapezoid is an isosceles trapezoid.

-If the diagonals of a trapezoid are congruent, the trapezoid is an isosceles trapezoid.

-If three or more parallel lines intercept congruent line segements on one transversal, then they intercept congruent line segments on any traversal.

Cyclic Quadrilaterals:

A cyclic quadrilateral is a quadrilateral that can be inscribed in a circle.
This means that the circle passes through all the four vertices of the quadrilateral.

Properties:

• The sum of the opposite angles of a cyclic quadrilateral is supplementary.

• The Perpendicular Bisectors of a quadrilateral are concurrent only if it is cyclic.

• The four perpendicular bisectors of the four sides meet at the center of the circle only if it is a cyclic quadrilateral.

• In a cyclic quadrilateral, the ratio of the diagonals equals the ratio of the sum of products of the sides that share the diagonal’s end points.

In a cyclic quadrilateral of sides a,b,c and d and of diagonals e and f:

$ac+bd=ef$

Proof:

Let $CD=e$

In triangles $CDE$ and $CDB$

$e^{2}=a^{2}+d^{2}-2ad\cos E=b^{2}+c^{2}-2bc\cos B$

But we know that:

Angles $E$ and $B$ are supplementary angles.

That means: $\cos B=-\cos E$

We get:

$\displaystyle{a^{2}+d^{2}-2ad\cos E}=b^{2}+c^{2}+2bc\cos E$

$\displaystyle{a^{2}+d^{2}-(b^{2}+c^{2})}=2bc\cos E+2ad\cos E$

$\displaystyle{(2bc+2ad)\cos E}=a^{2}+d^{2}-(b^{2}+c^{2})$

$\displaystyle{\cos E}=\frac{a^{2}+d^{2}-(b^{2}+c^{2})}{2bc+2ad}$

Now back to the diagonal:

$ <img src="https://www.mouctar.org/wp-content/ql-cache/quicklatex.com-53db70d41a364cdf6228be13fd6a1da8_l3.png" height="365" width="418" class="ql-img-displayed-equation quicklatex-auto-format" alt="\begin{align*}e^{2}&=a^{2}+d^{2}-2ad\frac{a^{2}+d^{2}-(b^{2}+c^{2})}{2bc+2ad}\\&=\frac{(a^{2}+d^{2})(2bc+2ad)-2ad((a^{2}+d^{2})-(b^{2}+c^{2}))}{2bc+2ad}\\&=\frac{(a^{2}+d^{2})(bc+ad)-ad((a^{2}+d^{2})-(b^{2}+c^{2}))}{bc+ad}\\&=\frac{(a^{2}+d^{2})(bc+ad-ad)+adb^{2}+adc^{2}}{bc+ad}\\&=\frac{(a^{2}+d^{2})bc+adb^{2}+adc^{2}}{bc+ad}\\&=\frac{ab\cdot ac+db\cdot dc+ab\cdot db+ac\cdot dc}{bc+ad}\\&=\frac{ac(ab+dc)+db(dc+ab)}{bc+ad}\\&=\frac{(ab+dc)(ac+db)}{bc+ad}\end{align*}" title="Rendered by QuickLaTeX.com"/>$

This yields:

$\displaystyle{e}=\sqrt{\frac{(ab+dc)(ac+db)}{bc+ad}}$

Now let’s calculate $EB=f$ :

In triangles $EDB$ and $ECB$

$f^{2}=a^{2}+b^{2}-2ab\cos D=d^{2}+c^{2}-2dc\cos C$

But we know that:

Angles $D$ and $C$ are supplementary angles.

That means: $\cos D=-\cos C$

We get:

$\displaystyle{a^{2}+b^{2}-2ab\cos D}=d^{2}+c^{2}+2dc\cos D$

$\displaystyle{a^{2}+b^{2}-(d^{2}+c^{2})}=2dc\cos E+2ab\cos E$

$\displaystyle{(2dc+2ab)\cos D}=a^{2}+b^{2}-(d^{2}+c^{2})$

$\displaystyle{\cos D}=\frac{a^{2}+b^{2}-(d^{2}+c^{2})}{2dc+2ab}$

Now back to the diagonal:

$ <img src="https://www.mouctar.org/wp-content/ql-cache/quicklatex.com-bdb8f3ce8abfdbacdd81c703bc732635_l3.png" height="365" width="418" class="ql-img-displayed-equation quicklatex-auto-format" alt="\begin{align*}f^{2}&=a^{2}+b^{2}-2ab\frac{a^{2}+b^{2}-(d^{2}+c^{2})}{2dc+2ab}\\&=\frac{(a^{2}+b^{2})(2dc+2ab)-2ab((a^{2}+b^{2})-(d^{2}+c^{2}))}{2dc+2ab}\\&=\frac{(a^{2}+b^{2})(dc+ab)-ab((a^{2}+b^{2})-(d^{2}+c^{2}))}{dc+ab}\\&=\frac{(a^{2}+b^{2})(dc+ab-ab)+abd^{2}+abc^{2}}{dc+ab}\\&=\frac{(a^{2}+b^{2})dc+abd^{2}+abc^{2}}{dc+ab}\\&=\frac{ad\cdot ac+db\cdot bc+ad\cdot db+ac\cdot bc}{dc+ab}\\&=\frac{ac(ad+bc)+db(bc+ad)}{dc+ab}\\&=\frac{(ad+bc)(ac+db)}{dc+ab}\end{align*}" title="Rendered by QuickLaTeX.com"/>$