Quadri-literacy

Definition 1. A quadrilateral is a four-sided shape.

For example, a rectangle is a quadrilateral whose internal angles are all $90^\circ$ .

Definition 2. A rhombus is a quadrilateral with equal side lengths.

Problem 1. Show that a quadrilateral is a square if and only if it is both a rectangle and a rhombus.

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Solution. $(\Leftarrow)$ Suppose a quadrilateral is both a rectangle and a rhombus.

As a rectangle, all of its angles are $90^\circ$ .
As a rhombus, all of its side lengths are equal.

Therefore, it must be a square.

$(\Rightarrow)$ Trivial.

Definition 3. A parallelogram is a quadrilateral with two pairs of parallel lines.

Problem 2. Show that the following are equivalent for a given quadrilateral $Q$ :

$Q$ is a parallelogram,
opposite sides in $Q$ are equal,
opposite angles in $Q$ are equal.

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Solution. $(1 \Rightarrow 2)$ Consider the parallelogram $Q = ABCD$ .

Draw the diagonal $BD$ . Since $AB \parallel DC$ , by alternate angles,

$\angle ABD = \angle CDB.$

As a common side, $BD = DB$ . Since $AD \parallel BC$ , by alternate angles,

$\angle ADB = \angle CBD.$

By the ASA Criterion, $\Delta ABD \equiv \Delta CDB$ . In particular, $AB = CD$ and $AD = CB$ , so that its opposite sides are equal.

$(2 \Rightarrow 3)$ Draw the diagonal and use the SSS Criterion to conclude that opposite angles equal each other.

$(3 \Rightarrow 1)$ Let $\alpha,\beta,\alpha,\beta$ denote the angles of the quadrilateral.

Since angles in a quadrilateral sum to $360^\circ$ ,

$2(\alpha + \beta) = 360^\circ \quad \Rightarrow \quad \alpha + \beta = 180^\circ.$

Since this equality holds for any pair of interior angles, the parallelogram must have two pairs of parallel sides.

Problem 3. Show that a rectangle is always a parallelogram. Furthermore, show that a parallelogram is a rectangle if and only if it has at least one interior right angle.

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Solution. Since all angles in a rectangle is $90^\circ$ , opposite pairs of angles are equal.

By Problem 2, a rectangle is a parallelogram.

$(\Leftarrow)$ Let $90^\circ, \alpha, \beta, \gamma$ denote the interior angles of the parallelogram, labelled anti-clockwise.

By Problem 2, $\alpha = \gamma$ and $\beta = 90^\circ$ . Since angles in a quadrilateral sum to $360^\circ$ ,

$\alpha + 90^\circ + \alpha + 90^\circ = 360^\circ \quad \Rightarrow \quad \alpha = 90^\circ.$

Therefore, all angles equal $90^\circ$ , and the parallelogram is a rectangle.

$(\Rightarrow)$ Trivial.

Problem 4. Show that a rhombus is always a parallelogram. Furthermore, show that a parallelogram is a rhombus if and only if it has at least one pair of equal adjacent sides.

(Click for Solution)

Solution. Since opposite sides in a rhombus are equal, by Problem 2, a rhombus is a parallelogram.

$(\Leftarrow)$ Let $w,x,y,z$ denote the sides of a parallelogram, labelled anti-clockwise.

By Problem 2, $w=y$ and $x=z$ . By hypothesis, suppose $w=x$ without loss of generality. Then $z=x=w=y$ . Therefore, all side lengths are equal, and the parallelogram is a rhombus.

$(\Rightarrow)$ Trivial.

Problem 5. Show that a rectangle is a square if and only if it has at least one pair of equal adjacent sides. Likewise, show that a rhombus is a square if and only if it has at least one interior right angle.

(Click for Solution)

Solution. We first prove the rectangle claim:

$(\Leftarrow)$ By Problem 3, a rectangle is a parallelogram. By hypothesis and Problem 4, it is a rhombus. By Problem 1, it is a square.
$(\Rightarrow)$ Trivial.

The rhombus claim follows similarly:

$(\Leftarrow)$ By Problem 4, a rhombus is a parallelogram. By hypothesis and Problem 2, it is a rectangle. By Problem 1, it is a square.
$(\Rightarrow)$ Trivial.

Definition 4. A trapezium is a quadrilateral with at least one pair of opposite sides that are parallel.

Problem 6. Show that a parallelogram is always a trapezium. Furthermore, show that a trapezium is a parallelogram if and only if it has at least one pair of equal opposite angles.

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Solution. Since a parallelogram has two pairs of equal and parallel sides, it has at least one pair of parallel sides, and is thus a trapezium.

$(\Leftarrow)$ Denote the angles of the trapezium by $\alpha, \beta, \alpha, \theta$ .

Given the pair of parallel sides, interior angles are supplementary, so that

$\alpha +\beta = 180^\circ = \alpha + \theta \quad \Rightarrow \quad \beta = \theta.$

By Problem 2, the trapezium is a parallelogram.

$(\Rightarrow)$ Trivial.

Definition 5. A kite is a quadrilateral with two pairs of adjacent sides that are equal in length.

Problem 7. Show that a kite has at least one pair of equal opposite angles and that its diagonals are perpendicular to each other.

(Click for Solution)

Solution. Consider the kite $ABCD$ below.

As base angles of isosceles triangles,

$\angle CBD = \angle CDB, \quad \angle ABD = \angle ADB.$

Hence,

$\begin{aligned} \angle ABC &= \angle ABD + \angle CBD \\ &= \angle ADB + \angle CDB \\ &= \angle ADC, \end{aligned}$

as required. Using the SAS Criterion, $\Delta ABC \equiv \Delta ADC$ . In particular, $\angle BAR = \angle DAR$ . As a kite, $AB = AD$ . As base angles of an isosceles triangle,

$\angle ABR = \angle ABD = \angle ADB = \angle ADR.$

Using the ASA Criterion, $\Delta ABR \equiv \angle ADR$ , so that $\angle ARB = \angle ARD$ . Since adjacent angles on a straight line are supplementary, $\angle ARB + \angle ARD = 180^\circ$ . Solving, $\angle ARB = 90^\circ$ .

Problem 8. Show that a quadrilateral is a rhombus if and only if it is both a kite and a trapezium.

(Click for Solution)

Solution. $(\Leftarrow)$ Suppose the quadrilateral is both a kite and a trapezium. By Problem 7, it has at least one pair of equal opposite angles. By Problem 6, it is a parallelogram. As a kite, it has at least one pair of equal adjacent sides. By Problem 4, it is a rhombus.

$(\Rightarrow)$ Trivial.

—Joel Kindiak, 24 Jan 26, 1651H

KindiakMath

Quadri-literacy

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Quadri-literacy

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