Construction of Quadrilaterals

Construction of a General Quadrilateral (Given specific side and angle combinations)

A quadrilateral is a polygon with four sides and four vertices. Unlike triangles, which are rigid when side lengths are fixed, quadrilaterals can be flexible (e.g., a square can be deformed into a rhombus without changing side lengths). Therefore, knowing only the lengths of the four sides is not sufficient to construct a unique quadrilateral.

To construct a unique quadrilateral, we typically need a minimum of five independent measurements. These measurements can be combinations of side lengths, angle measures, and diagonal lengths. The general strategy for constructing any quadrilateral is to divide it into two triangles using a diagonal and then construct these triangles using the appropriate congruence criteria (SSS, SAS, ASA, AAS, RHS) based on the given information.

Common Combinations for Unique Construction

Here are some standard sets of five measurements that uniquely determine a quadrilateral (up to congruence) and how they are typically constructed:

• Four sides and one diagonal (SSSS D): Example: $AB, BC, CD, DA$ and diagonal $AC$.
• Four sides and one angle (SSAS S - variant): Example: $AB, BC, CD, DA$ and angle $\angle B$.
• Three sides and two included angles (SASAS): Example: $AB, BC, CD$ and angles $\angle B, \angle C$.
• Two adjacent sides and three angles (ASASA - variant): Example: $AB, BC$ and angles $\angle A, \angle B, \angle C$. (Note: if three angles are given, the fourth is determined as their sum must be $360^\circ$).

Construction Example: Given Four Sides and One Diagonal (SSSS D)

This is a direct application of the SSS triangle construction.

Given: The lengths of the four sides of a quadrilateral $ABCD$: $AB=a, BC=b, CD=c, DA=d$, and the length of one diagonal, $AC=e$.

Tools: Ruler, Compass, Straightedge.

Goal: Construct the quadrilateral $ABCD$.

Steps:

1. Construct the First Triangle: Use the diagonal $AC$ as a base and the two sides adjacent to it (say $AB$ and $BC$) to construct the first triangle, $\triangle ABC$, using the SSS criterion (I1).
   • Draw the line segment $AC$ of length $e$.
   • With $A$ as the center and compass width equal to the length of side $AB$ ($a$), draw an arc on one side of $AC$.
   • With $C$ as the center and compass width equal to the length of side $BC$ ($b$), draw another arc intersecting the first arc. Label the intersection point $B$.
   • Use the straightedge to draw the line segments $AB$ and $CB$.
2. Construct the Second Triangle: Using the same diagonal $AC$ as a base and the other two sides ($AD$ and $CD$), construct the second triangle, $\triangle ADC$, on the opposite side of $AC$ from where $B$ was constructed, again using the SSS criterion (I1).
   • Use the existing segment $AC = e$.
   • With $A$ as the center and compass width equal to the length of side $DA$ ($d$), draw an arc on the opposite side of $AC$ from point $B$.
   • With $C$ as the center and compass width equal to the length of side $CD$ ($c$), draw another arc intersecting the arc drawn from $A$. Label the intersection point $D$.
   • Use the straightedge to draw the line segments $AD$ and $CD$.
3. Result: The figure formed by the vertices $A, B, C, D$ in that order is the required quadrilateral $ABCD$.

Justification: By construction, $\triangle ABC$ has sides $AB=a, BC=b, AC=e$ and $\triangle ADC$ has sides $CD=c, DA=d, AC=e$. These match the given side and diagonal lengths. Since two triangles are uniquely determined by their side lengths (provided the triangle inequality holds for each triangle), the positions of $B$ and $D$ relative to $AC$ are fixed (up to reflection across $AC$). Constructing $B$ and $D$ on opposite sides of $AC$ ensures the formation of a quadrilateral. Thus, the quadrilateral $ABCD$ is uniquely determined.

Construction Example: Given Three Sides and Two Included Angles (SASAS)

This construction combines drawing sides and constructing angles.

Given: The lengths of three consecutive sides of a quadrilateral $ABCD$: $AB=a, BC=b, CD=c$, and the measures of the two angles included between these sides, $\angle B = \beta$ and $\angle C = \gamma$.

Tools: Ruler, Compass, Straightedge, Protractor (or methods for specific constructible angles).

Goal: Construct the quadrilateral $ABCD$.

Steps:

1. Draw the Middle Side: Use the ruler to draw the line segment $BC$ of length $b$.
2. Construct the First Included Angle and Side: At vertex $B$, construct the given angle $\beta$. Draw a ray $BX$ starting from $B$ such that $\angle CBX = \beta$. On this ray $BX$, use the ruler or compass to measure and mark off the length of side $AB$, which is $a$. Label the endpoint as $A$. So, $BA = a$.
3. Construct the Second Included Angle and Side: At vertex $C$, construct the given angle $\gamma$. Draw a ray $CY$ starting from $C$ such that $\angle BCY = \gamma$. On this ray $CY$, use the ruler or compass to measure and mark off the length of side $CD$, which is $c$. Label the endpoint as $D$. So, $CD = c$.
4. Join the Remaining Vertices: The four vertices $A, B, C, D$ are now located. Use the straightedge to draw the line segment connecting point $A$ to point $D$.
5. Result: The figure $ABCD$ is the required quadrilateral.

Justification: By construction, $BC=b$, $\angle ABC = \beta$, $\angle BCD = \gamma$, $AB=a$, and $CD=c$. These match the given conditions. The positions of $A$ relative to $B$ and $C$ relative to $D$ are fixed by the side lengths and angles. Once $A, B, C, D$ are located, the quadrilateral is uniquely determined.

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Conditions for Unique Construction of a Quadrilateral

A quadrilateral is determined by its shape and size. As highlighted earlier, simply knowing the lengths of the four sides is insufficient to fix the shape (e.g., a square and a rhombus with the same side length). To ensure a unique quadrilateral is constructed, we need enough constraints on the vertices' relative positions. This generally translates to needing a specific number of independent measurements.

The Requirement for Five Independent Measurements

A triangle has 3 vertices, and its shape is determined by 3 measurements (SSS, SAS, ASA). A quadrilateral has 4 vertices. To uniquely define the position of 4 vertices in a plane relative to each other, you need more information than just side lengths. Each vertex's position relative to a fixed coordinate system requires 2 coordinates (say x, y). So, 4 vertices might seem to require $4 \times 2 = 8$ pieces of information. However, fixing the first vertex (say at (0,0)) removes 2 degrees of freedom. Fixing the second vertex along an axis (say on the x-axis at (a,0)) removes 2 more. The remaining 2 vertices need to be located relative to these, requiring 4 more pieces of information. Total: $2+2+4=8$. However, since the figure can be moved (translation) and rotated freely, this reduces the required independent measurements. Fixing two points in space removes 4 degrees of freedom (2 for position, 2 for orientation). To locate the third point relative to these two fixed points requires 2 more measurements (e.g., distances from the two points). To locate the fourth point relative to the first three requires 3 measurements (e.g., distances from three points), but one of these might be redundant if it results in a non-planar figure or over-constrains a planar figure. A more intuitive approach for planar quadrilaterals is that they can be divided into two triangles by a diagonal. A triangle needs 3 measurements. A quadrilateral needs the 3 measurements for the first triangle, plus 3 measurements for the second triangle, but one side (the shared diagonal) is counted twice. So, $3+3-1 = 5$ independent measurements are typically needed to uniquely define a planar quadrilateral.

The five measurements must be chosen carefully to effectively fix the angles or the diagonal lengths relative to the sides.

Combinations that Yield a Unique Quadrilateral

Here are the standard sets of five measurements that uniquely determine a quadrilateral up to congruence, allowing construction using basic triangle methods:

1. Four sides and one diagonal: e.g., $AB, BC, CD, DA, AC$. This breaks the quadrilateral into $\triangle ABC$ and $\triangle ADC$, both constructible by SSS.
2. Four sides and one angle: e.g., $AB, BC, CD, DA, \angle B$. Construct $\triangle ABC$ using SAS ($AB, \angle B, BC$). Then, with $AC$ determined, construct $\triangle ADC$ using SSS ($AD, CD, AC$).
3. Three sides and two included angles: e.g., $AB, BC, CD, \angle B, \angle C$. Construct $\triangle ABC$ using SAS ($AB, \angle B, BC$). Then construct $\triangle BCD$ using SAS ($BC, \angle C, CD$). Join $AD$. This is slightly misleading; the construction typically proceeds as shown in I1 SASAS example: Draw $BC$, construct $\angle B$ and $AB$, construct $\angle C$ and $CD$, join $AD$.
4. Two adjacent sides and three angles: e.g., $AB, BC, \angle A, \angle B, \angle C$. Since the sum of angles is $360^\circ$, the fourth angle $\angle D = 360^\circ - (\angle A + \angle B + \angle C)$ is determined. You can then use combinations of ASA/AAS. For example, draw $AB$, construct $\angle A$ and $\angle B$. This determines the directions of $AD$ and $BC$. Mark $BC$ length on the $\angle B$ ray. From $C$, construct angle $\angle C$. Its intersection with the $\angle A$ ray gives $D$.

Other combinations of five measurements might also work, provided they uniquely fix the figure (e.g., two adjacent sides, one opposite side, and two angles). However, the five combinations above are the most commonly taught and used for construction.

Combinations that Do Not Guarantee Uniqueness

• Four sides only: As discussed, quadrilaterals with the same side lengths can have different shapes (e.g., rhombus vs. square, or various parallelograms with fixed side lengths).
• Four angles only: This only determines the *shape* (similar quadrilaterals), not the *size*. You need at least one side length to fix the scale. Even with one side, the shape isn't unique (e.g., infinitely many rectangles exist with angles 90-90-90-90 and a fixed side length).
• Three sides and two diagonals: e.g., $AB, BC, CD, AC, BD$. This can sometimes lead to two possible quadrilaterals (reflections of each other across a diagonal or side, or different arrangements).

Construction of Parallelograms

A parallelogram is a special type of quadrilateral with specific properties:

• Opposite sides are parallel ($\textsf{AB} \parallel \textsf{CD}$, $\textsf{AD} \parallel \textsf{BC}$) and equal ($\textsf{AB = CD}$, $\textsf{AD = BC}$).
• Opposite angles are equal ($\angle \textsf{A} = \angle \textsf{C}$, $\angle \textsf{B} = \angle \textsf{D}$).
• Adjacent angles are supplementary ($\angle \textsf{A} + \angle \textsf{B} = 180^\circ$, etc.).
• Diagonals bisect each other.

Due to these properties, fewer than five measurements are often needed to construct a unique parallelogram. For instance, two adjacent sides and the included angle are sufficient (related to SAS). Three pieces of information are often enough if they are chosen appropriately.

Common Parallelogram Constructions

1. Given two adjacent sides and the included angle

This is analogous to the SAS construction for a triangle, but we use the parallelogram properties to find the fourth vertex.

Given: Lengths of two adjacent sides, say $AB=a$ and $AD=b$, and the measure of the included angle $\angle A = \alpha$.

Tools: Ruler, Compass, Straightedge, Protractor (or methods for specific angles).

Goal: Construct parallelogram $ABCD$.

Steps:

1. Draw First Side: Draw the line segment $AB$ of length $a$.
2. Construct Included Angle and Second Side: At vertex $A$, construct the given angle $\alpha$. Draw a ray $AX$ such that $\angle BAX = \alpha$. On this ray $AX$, measure and mark off the length of side $AD$, which is $b$. Label the endpoint as $D$. So, $AD = b$.
3. Locate Fourth Vertex using Arcs: We know that opposite sides of a parallelogram are equal. $BC$ must be equal to $AD=b$, and $CD$ must be equal to $AB=a$.
   • With $B$ as the center and compass width equal to $b$ (length $AD$), draw an arc. Point $C$ must be on this arc.
   • With $D$ as the center and compass width equal to $a$ (length $AB$), draw another arc that intersects the arc drawn from $B$. Label the intersection point $C$. Point $C$ must be on this arc.
   The intersection $C$ is the unique location for the fourth vertex.
4. Complete the Parallelogram: Use the straightedge to draw the line segments $BC$ and $DC$.

The figure $ABCD$ is the required parallelogram.

Alternative Method (Using Parallel Lines):

1. Draw side $AB=a$.
2. At $A$, construct $\angle A=\alpha$ and draw ray $AD$ such that $AD=b$.
3. Draw a line through $D$ parallel to $AB$ (I1).
4. Draw a line through $B$ parallel to $AD$ (I1).
5. The intersection of the parallel lines from $D$ and $B$ is vertex $C$.

Justification: By construction (first method), $AB=a, AD=b, \angle A = \alpha, BC=b, CD=a$. Since opposite sides are equal ($AB=CD=a$ and $AD=BC=b$), the quadrilateral $ABCD$ is a parallelogram. In the alternative method, $AB \parallel DC$ and $AD \parallel BC$ by construction of parallel lines, which is the definition of a parallelogram.

2. Given two adjacent sides and one diagonal

This is a direct application of the SSS triangle construction.

Given: Lengths of two adjacent sides, say $AB=a$ and $BC=b$, and the length of one diagonal, say $AC=d$.

Tools: Ruler, Compass, Straightedge.

Goal: Construct parallelogram $ABCD$.

Steps:

1. Construct First Triangle: Construct $\triangle ABC$ using the SSS criterion (I1) with sides $AB=a, BC=b,$ and $AC=d$.
2. Locate Fourth Vertex using Arcs: We know $AD = BC = b$ and $CD = AB = a$.
   • With $A$ as the center and compass width $b$ (length $BC$), draw an arc. Point $D$ must be on this arc.
   • With $C$ as the center and compass width $a$ (length $AB$), draw another arc intersecting the arc drawn from $A$. Label the intersection point $D$. Point $D$ must be on this arc. Ensure $D$ is on the opposite side of $AC$ from $B$.
3. Complete the Parallelogram: Join $AD$ and $CD$ using a straightedge.

The figure $ABCD$ is the required parallelogram.

Justification: By construction, $AB=a, BC=b, AC=d, AD=b, CD=a$. Since $AB=CD$ and $BC=AD$, the quadrilateral $ABCD$ has opposite sides equal, which is a property of a parallelogram. Therefore, $ABCD$ is a parallelogram.

3. Given diagonals and the angle between them

Given: Lengths of the two diagonals, say $AC=d_1$ and $BD=d_2$, and the measure of the angle between them, say $\theta$.

Tools: Ruler, Compass, Straightedge, Protractor (or methods for specific angles).

Goal: Construct parallelogram $ABCD$.

Property Used: Diagonals of a parallelogram bisect each other. Let the intersection point be $O$. Then $AO = OC = d_1/2$ and $BO = OD = d_2/2$. The angle between $AC$ and $BD$ is $\theta$.

Steps:

1. Draw First Diagonal: Draw the line segment $AC$ of length $d_1$.
2. Find Midpoint and Construct Angle: Find the midpoint $O$ of $AC$ (using perpendicular bisector construction or by measuring $d_1/2$). At point $O$, construct an angle equal to $\theta$. Draw a line $L_1$ passing through $O$ such that it makes an angle $\theta$ with $AC$.
3. Construct Supplementary Angle (Optional): Construct a line $L_2$ through $O$ perpendicular to $L_1$. This line will make an angle of $180^\circ - \theta$ with $AC$. Or simply extend $L_1$ through $O$.
4. Mark Half Diagonals: On the line $L_1$ (and its extension through $O$), measure and mark points $B$ and $D$ such that $OB = OD = d_2/2$. Ensure $B$ and $D$ are on opposite sides of $AC$.
5. Complete the Parallelogram: Join the points $A$ to $B$, $B$ to $C$, $C$ to $D$, and $D$ to $A$ using a straightedge.

The figure $ABCD$ is the required parallelogram.

Justification: By construction, $AC$ and $BD$ are line segments of lengths $d_1$ and $d_2$ respectively. They intersect at $O$, and $O$ is the midpoint of both $AC$ ($AO=OC=d_1/2$) and $BD$ ($BO=OD=d_2/2$). The angle between the diagonals at $O$ is $\theta$ (or $180-\theta$). Since the diagonals of quadrilateral $ABCD$ bisect each other, $ABCD$ is a parallelogram. The given diagonal lengths and the angle between them are incorporated into the construction.

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Construction of Rectangles

A rectangle is a special type of parallelogram where all four interior angles are right angles ($90^\circ$). As a parallelogram, its opposite sides are equal in length and parallel. To construct a unique rectangle, knowing the lengths of its adjacent sides (length and breadth) is sufficient. Alternatively, knowing one side and a diagonal, or both diagonals and the angle between them (which must be $90^\circ$ for the diagonals to be equal) can also define a unique rectangle.

Construction: Given Length and Breadth

This is the most common method, using the definitions of sides and angles.

Given: The length ($l$) and the breadth ($b$) of the rectangle. Let's construct rectangle $ABCD$ where $AB=l$ and $BC=b$.

Tools: Ruler, Compass, Straightedge.

Goal: Construct rectangle $ABCD$.

Steps:

1. Draw the Base: Use the ruler to draw the line segment $AB$ of length $l$.
2. Construct a Right Angle: At vertex $A$, construct a perpendicular line (a $90^\circ$ angle) using the compass and straightedge method (I1 - Perpendicular at a Point on the Line). Draw a ray $AX$ starting from $A$ such that $\angle BAX = 90^\circ$. This ray will contain the side $AD$.
3. Mark the Breadth: The side $AD$ must have length equal to the breadth $b$. Set the compass width equal to $b$. Place the pointed end of the compass on $A$ and draw an arc that intersects the ray $AX$ drawn in Step 2. Label the point of intersection $D$. So, $AD=b$.
4. Locate the Fourth Vertex (C) using Arcs: We know that in a rectangle, opposite sides are equal, so $BC$ must be equal to $AD = b$, and $CD$ must be equal to $AB = l$.
   • With $D$ as the center and compass width equal to $l$ (length of $AB$), draw an arc. Point $C$ must be on this arc.
   • With $B$ as the center and compass width equal to $b$ (length of $AD$ or $BC$), draw another arc intersecting the arc drawn from $D$. Label the intersection point $C$. Point $C$ must be on this arc.
   The intersection point $C$ is the unique location for the fourth vertex.
5. Complete the Rectangle: Use the straightedge to draw the line segments connecting point $B$ to point $C$ and point $D$ to point $C$.

The resulting figure $ABCD$ is the required rectangle.

Alternative Method (Using Parallel Lines):

1. Draw $AB=l$.
2. At $A$, construct a $90^\circ$ angle and draw ray $AD$ such that $AD=b$.
3. At $B$, construct a $90^\circ$ angle and draw ray $BC'$ such that $BC'=b$.
4. Join $D$ and $C'$. The point $C'$ is actually $C$. $ABCD$ is the rectangle. (Or draw a line through $D$ parallel to $AB$ and a line through $B$ parallel to $AD$; their intersection is $C$).

Justification: By construction (first method), $AB=l, AD=b, \angle A = 90^\circ$. We constructed $C$ such that $BC=b$ and $DC=l$. Since $AB=DC=l$ and $AD=BC=b$, the quadrilateral has opposite sides equal, making it a parallelogram. Since one angle is $90^\circ$, and it's a parallelogram, all angles are $90^\circ$. Thus, $ABCD$ is a rectangle with length $l$ and breadth $b$.

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Construction of Rhombi

A rhombus is a special type of parallelogram where all four sides are equal in length. Key properties of a rhombus include:

• All sides are equal.
• Opposite angles are equal.
• Adjacent angles are supplementary.
• Diagonals bisect each other at right angles.
• Diagonals bisect the angles of the rhombus.

To construct a unique rhombus, you need fewer measurements than a general quadrilateral, typically two. Common scenarios involve given side length and an angle, or the lengths of the two diagonals.

Construction 1: Given Side and One Angle

This is a direct application of the parallelogram construction given two adjacent sides and the included angle, simplified because the side lengths are equal.

Given: The side length $s$ and the measure of one interior angle $\alpha$. Let's construct rhombus $ABCD$ with $AB=s$ and $\angle A = \alpha$.

Tools: Ruler, Compass, Straightedge, Protractor (or methods for specific constructible angles).

Goal: Construct rhombus $ABCD$.

Steps:

1. Construct Angle and First Two Sides: Draw the line segment $AB$ of length $s$. At vertex $A$, construct the given angle $\alpha$. Draw a ray $AX$ such that $\angle BAX = \alpha$. On this ray $AX$, measure and mark off the length of side $AD$, which must also be $s$. Label the endpoint as $D$. So, $AD=s$.
2. Locate the Fourth Vertex (C) using Arcs: We know that all four sides of a rhombus are equal to $s$.
   • With $B$ as the center and compass width equal to $s$, draw an arc. Point $C$ must be on this arc.
   • With $D$ as the center and compass width equal to $s$, draw another arc that intersects the arc drawn from $B$. Label the intersection point $C$. Point $C$ must be on this arc.
   The intersection point $C$ is the unique location for the fourth vertex.
3. Complete the Rhombus: Use the straightedge to draw the line segments connecting point $B$ to point $C$ and point $D$ to point $C$.

The figure $ABCD$ is the required rhombus.

Justification: By construction, $AB=s, AD=s, \angle A = \alpha, BC=s, CD=s$. Since all four sides are equal ($AB=BC=CD=DA=s$), the quadrilateral $ABCD$ is a rhombus. The angle $\angle A$ is equal to the given angle $\alpha$.

Construction 2: Given Diagonals

This construction uses the property that the diagonals of a rhombus are perpendicular bisectors of each other.

Given: The lengths of the two diagonals $d_1$ and $d_2$. Let the diagonals be $AC$ and $BD$ with lengths $d_1$ and $d_2$ respectively.

Tools: Ruler, Compass, Straightedge.

Goal: Construct rhombus $ABCD$.

Steps:

1. Draw First Diagonal and Perpendicular Bisector: Draw one of the diagonals, say $AC$, with length $d_1$. Construct the perpendicular bisector of the segment $AC$ (I3). Label the intersection point of the bisector and $AC$ as $O$. This point $O$ is the midpoint of $AC$. The perpendicular bisector is the line containing the other diagonal $BD$.
2. Mark Half of Second Diagonal: Calculate half the length of the second diagonal, $d_2/2$.
3. Locate the Other Two Vertices (B and D): The second diagonal $BD$ lies on the perpendicular bisector constructed in Step 1, and its midpoint is $O$. Place the pointed end of the compass on $O$. Set the compass width equal to $d_2/2$. Draw an arc that intersects the perpendicular bisector above $AC$. Label the intersection point $B$. Draw another arc with the same radius $d_2/2$ intersecting the perpendicular bisector below $AC$. Label this intersection point $D$. So, $OB = OD = d_2/2$.
4. Complete the Rhombus: Use the straightedge to draw the line segments connecting the vertices in order: $A$ to $B$, $B$ to $C$, $C$ to $D$, and $D$ to $A$.

The figure $ABCD$ is the required rhombus.

Justification: By construction, the diagonals $AC$ and $BD$ intersect at $O$, which is the midpoint of both segments ($AO=OC=d_1/2$, $BO=OD=d_2/2$). Also, the diagonals are perpendicular at $O$ because $BD$ lies on the perpendicular bisector of $AC$. A quadrilateral whose diagonals are perpendicular bisectors of each other is a rhombus. Thus, $ABCD$ is a rhombus with diagonals of lengths $d_1$ and $d_2$.

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Construction of Squares

A square is the most specific type of parallelogram. It is a rectangle with all sides equal, or equivalently, a rhombus with all angles equal (hence, right angles). Key properties:

• All four sides are equal.
• All four angles are right angles ($90^\circ$).
• Opposite sides are parallel.
• Diagonals are equal in length, bisect each other at right angles, and bisect the angles (forming $45^\circ$ angles).

Because of its highly constrained nature, a square is uniquely determined by just one measurement: either the side length or the diagonal length.

Construction 1: Given Side Length

This construction is identical to constructing a rectangle where length equals breadth, or constructing a rhombus with a $90^\circ$ angle.

Given: The side length $s$. Let's construct square $ABCD$ with side length $s$.

Tools: Ruler, Compass, Straightedge.

Goal: Construct square $ABCD$.

Steps:

1. Draw First Side: Use the ruler to draw the line segment $AB$ of length $s$.
2. Construct a Right Angle: At vertex $A$, construct a perpendicular line (a $90^\circ$ angle) using the compass and straightedge method (I1 - Perpendicular at a Point on the Line). Draw a ray $AX$ starting from $A$ such that $\angle BAX = 90^\circ$. This ray will contain the side $AD$.
3. Mark the Side Length: The side $AD$ must also have length $s$. Set the compass width equal to $s$. Place the pointed end of the compass on $A$ and draw an arc that intersects the ray $AX$. Label the point of intersection $D$. So, $AD=s$.
4. Locate the Fourth Vertex (C) using Arcs: We know that all four sides of a square are equal to $s$.
   • With $D$ as the center and compass width equal to $s$, draw an arc. Point $C$ must be on this arc.
   • With $B$ as the center and compass width equal to $s$, draw another arc that intersects the arc drawn from $D$. Label the intersection point $C$. Point $C$ must be on this arc.
   The intersection point $C$ is the unique location for the fourth vertex.
5. Complete the Square: Use the straightedge to draw the line segments connecting point $B$ to point $C$ and point $D$ to point $C$.

The resulting figure $ABCD$ is the required square.

Justification: By construction, $AB=s, AD=s, \angle A = 90^\circ$. We constructed $C$ such that $BC=s$ and $CD=s$. Since all four sides are equal ($AB=BC=CD=DA=s$), the quadrilateral is a rhombus. Since one angle $\angle A = 90^\circ$, and it's a rhombus, all angles are $90^\circ$. Thus, $ABCD$ is a square with side length $s$.

Construction 2: Given Diagonal Length

This construction uses the property that the diagonals of a square are equal in length, bisect each other at right angles, and their intersection is the center of the square.

Given: The length of a diagonal $d$. Let the diagonal be $AC$ with length $d$.

Tools: Ruler, Compass, Straightedge.

Goal: Construct square $ABCD$.

Steps:

1. Draw First Diagonal and Perpendicular Bisector: Draw one of the diagonals, say $AC$, with length $d$. Construct the perpendicular bisector of the segment $AC$ (I3). Label the intersection point of the bisector and $AC$ as $O$. This point $O$ is the midpoint of $AC$. The perpendicular bisector is the line containing the other diagonal $BD$.
2. Mark Half of Diagonal Length: The diagonals of a square are equal and bisect each other at right angles. So, the length of the other diagonal $BD$ is also $d$, and the point $O$ is the midpoint of $BD$ as well. Calculate half the length of the diagonal: $d/2$. Note that in a square, $AO=OC=BO=OD=d/2$.
3. Locate the Other Two Vertices (B and D): The second diagonal $BD$ lies on the perpendicular bisector constructed in Step 1, its midpoint is $O$, and its length is $d$. Place the pointed end of the compass on $O$. Set the compass width equal to $d/2$. Draw an arc that intersects the perpendicular bisector above $AC$. Label the intersection point $B$. Draw another arc with the same radius $d/2$ intersecting the perpendicular bisector below $AC$. Label this intersection point $D$. So, $OB = OD = d/2$.
4. Complete the Square: Use the straightedge to draw the line segments connecting the vertices in order: $A$ to $B$, $B$ to $C$, $C$ to $D$, and $D$ to $A$.

The figure $ABCD$ is the required square.

Justification: By construction, the diagonals $AC$ and $BD$ intersect at $O$. $AO=OC=d/2$ and $BO=OD=d/2$. Also, the diagonals are perpendicular at $O$ because $BD$ lies on the perpendicular bisector of $AC$. Since $AO=OC=BO=OD$, the diagonals are equal ($AC=BD=d$) and bisect each other at right angles. A quadrilateral with equal diagonals that bisect each other at right angles is a square. Thus, $ABCD$ is a square with diagonal length $d$.

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