Construction of Basic Triangles

Construction of a Triangle given Three Sides (SSS Criterion)

Constructing a triangle given the lengths of its three sides is a fundamental geometric construction. This is based on the SSS (Side-Side-Side) congruence criterion, which states that if three sides of one triangle are congruent to three sides of another triangle, then the triangles are congruent. This implies that a triangle is uniquely determined by the lengths of its three sides.

Condition for Construction

Before attempting construction, it is crucial to check if a triangle with the given side lengths is actually possible. The condition is given by the Triangle Inequality Theorem:

The sum of the lengths of any two sides of a triangle must be strictly greater than the length of the third side.

If the given side lengths are $a$, $b$, and $c$, a triangle can be constructed if and only if all three of these inequalities hold:

$a + b > c$

... (1)

$a + c > b$

... (2)

$b + c > a$

... (3)

If any of these conditions is not met (i.e., the sum is less than or equal to the third side), the arcs in the construction steps will not intersect, and a triangle cannot be formed with the given side lengths.

Construction Steps

Given: The lengths of the three sides of a triangle, say $a$, $b$, and $c$. We aim to construct $\triangle ABC$ such that the side opposite vertex $A$ has length $a$ ($BC=a$), the side opposite vertex $B$ has length $b$ ($AC=b$), and the side opposite vertex $C$ has length $c$ ($AB=c$).

Tools: Ruler (for measuring lengths), Compass (for drawing arcs), Straightedge (for drawing line segments).

Goal: Construct $\triangle ABC$ with given side lengths $a, b, c$.

Steps:

1. Draw the Base Side: Use the ruler to draw one of the sides as the base of the triangle. It's usually convenient to draw the longest side as the base, but any side will work. For example, draw the line segment $BC$ with length equal to $a$.
2. Draw Arc from One Endpoint: Place the pointed end of the compass on vertex $B$. Set the compass width equal to the length of the side $AB$, which is given as $c$. Draw an arc above (or below) the base line segment $BC$. The third vertex, $A$, must lie somewhere on this arc, because it is $c$ units away from $B$.
3. Draw Arc from the Other Endpoint: Now, place the pointed end of the compass on vertex $C$. Set the compass width equal to the length of the side $AC$, which is given as $b$. Draw another arc above (or below) the base line segment $BC$ so that it intersects the arc drawn in Step 2. The third vertex, $A$, must also lie somewhere on this second arc, because it is $b$ units away from $C$. The intersection of the two arcs is the only point that is $c$ units from $B$ and $b$ units from $C$ simultaneously. Label the point of intersection of the two arcs as $A$.
4. Join the Vertices: Use the straightedge to draw the line segments connecting point $A$ to point $B$ and point $A$ to point $C$.

The resulting figure is $\triangle ABC$. This triangle has sides $BC$ of length $a$ (by construction), $AB$ of length $c$ (radius of the arc from B), and $AC$ of length $b$ (radius of the arc from C). Thus, it is the required triangle with the given side lengths.

Justification:

By the steps of construction, the base $BC$ is drawn with length $a$. Point $A$ is located such that its distance from $B$ is $c$ (since it lies on the arc with radius $c$ centered at $B$) and its distance from $C$ is $b$ (since it lies on the arc with radius $b$ centered at $C$). Therefore, $AB = c$ and $AC = b$. The triangle $ABC$ constructed thus has side lengths $a, b, c$ as required. Provided the triangle inequality holds, these three lengths uniquely determine the triangle (SSS congruence criterion). If the inequality does not hold, the arcs will not intersect, indicating that such a triangle cannot exist.

Example

Answer:

Construction of a Triangle given Two Sides and the Included Angle (SAS Criterion)

Constructing a triangle given the lengths of two sides and the measure of the angle formed by these two sides (the included angle) is another basic construction. This is based on the SAS (Side-Angle-Side) congruence criterion, which states that if two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then the triangles are congruent. This criterion ensures that the triangle is uniquely determined by these three pieces of information.

Construction Steps

Given: The lengths of two sides, say $b$ and $c$, and the measure of the angle included between them, say $\angle A$. We aim to construct $\triangle ABC$ such that $AB = c$, $AC = b$, and $\angle BAC = \angle A$.

Tools: Ruler (for measuring lengths), Compass (for drawing arcs), Straightedge (for drawing line segments and rays), Protractor (for measuring and constructing angles, or compass/straightedge methods for specific constructible angles).

Goal: Construct $\triangle ABC$ with sides $b, c$ and included angle $\angle A$.

Steps:

1. Draw One of the Given Sides: Use the ruler to draw one of the two given sides. For example, draw the line segment $AB$ with length equal to $c$. This segment will be one side of the angle $\angle A$.
2. Construct the Included Angle: At vertex $A$ (one endpoint of the segment drawn in Step 1), construct the given angle $\angle A$. Draw a ray $AX$ starting from $A$ such that $\angle BAX$ is equal to the given measure of $\angle A$. If $\angle A$ is a standard constructible angle (like $60^\circ, 90^\circ, 45^\circ, \dots$), use compass and straightedge. Otherwise, use a protractor for measurement. This ray $AX$ will contain the other side of the included angle.
3. Mark the Second Side Length: The second given side is $AC$, with length $b$. Place the pointed end of the compass on vertex $A$. Set the compass width equal to the length $b$. Draw an arc that intersects the ray $AX$ drawn in Step 2. The point of intersection is vertex $C$, as it must be $b$ units away from $A$ along the arm of the angle $\angle A$. Label the intersection point as $C$.
4. Join the Vertices: Use the straightedge to draw the line segment connecting point $B$ to point $C$.

The resulting figure is $\triangle ABC$. By construction, side $AB$ has length $c$, side $AC$ has length $b$, and the included angle $\angle BAC$ is equal to the given angle $\angle A$. Thus, this is the required triangle.

Justification:

The construction directly creates a triangle with sides $AB=c$ and $AC=b$, and the angle $\angle BAC$ between them is equal to the given $\angle A$. The location of vertex $C$ is uniquely determined by its distance $b$ from $A$ along the ray $AX$. Once $A$, $B$, and $C$ are fixed, the triangle is formed. By the SAS congruence criterion, a triangle is uniquely determined by two sides and the included angle. Therefore, the construction accurately produces the required triangle.

Example

Answer:

Construction of a Triangle given Two Angles and the Included Side (ASA Criterion)

Constructing a triangle when the measures of two angles and the length of the side located between these two angles (the included side) are known is a third basic construction. This is based on the ASA (Angle-Side-Angle) congruence criterion, which states that if two angles and the included side of one triangle are congruent to two angles and the included side of another triangle, then the triangles are congruent. This criterion ensures the unique determination of the triangle by these three pieces of information.

Condition for Construction

For any triangle to exist, the sum of its interior angles must be $180^\circ$. If you are given two angles, say $\angle B$ and $\angle C$, their sum must be strictly less than $180^\circ$ for the third angle ($\angle A = 180^\circ - (\angle B + \angle C)$) to be positive. If $\angle B + \angle C \ge 180^\circ$, a triangle cannot be formed.

Construction Steps

Given: The measures of two angles, say $\angle B$ and $\angle C$, and the length of the side included between them, say $a$ (side $BC$). We aim to construct $\triangle ABC$ such that $BC = a$, $\angle ABC = \angle B$, and $\angle ACB = \angle C$.

Tools: Ruler (for measuring length), Compass (for drawing arcs), Straightedge (for drawing line segments and rays), Protractor (for measuring and constructing angles, or compass/straightedge methods for specific constructible angles).

Goal: Construct $\triangle ABC$ with angles $\angle B, \angle C$ and included side $a=BC$.

Steps:

1. Draw the Included Side: Use the ruler to draw the line segment that is the included side. Draw the line segment $BC$ with length equal to $a$. This segment will form one arm of both angles $\angle B$ and $\angle C$.
2. Construct the First Angle: At vertex $B$ (one endpoint of the segment drawn in Step 1), construct the given angle $\angle B$. Draw a ray $BX$ starting from $B$ such that $\angle CBX$ is equal to the given measure of $\angle B$. This ray $BX$ will contain the side $AB$ of the triangle.
3. Construct the Second Angle: At vertex $C$ (the other endpoint of the segment drawn in Step 1), construct the given angle $\angle C$. Draw a ray $CY$ starting from $C$ such that $\angle BCY$ is equal to the given measure of $\angle C$. This ray $CY$ will contain the side $AC$ of the triangle. Make sure to draw the ray $CY$ on the same side of $BC$ as ray $BX$.
4. Locate the Third Vertex: The third vertex of the triangle, $A$, must lie on both ray $BX$ and ray $CY$. Therefore, the point $A$ is the point of intersection of rays $BX$ and $CY$. Extend the rays $BX$ and $CY$ until they intersect. Label the point of intersection as $A$.

The resulting figure is $\triangle ABC$. By construction, side $BC$ has length $a$, angle $\angle ABC$ (which is $\angle CBX$) is equal to $\angle B$, and angle $\angle ACB$ (which is $\angle BCY$) is equal to $\angle C$. Thus, this is the required triangle.

Justification:

The construction directly produces a line segment $BC$ of length $a$ and angles $\angle B$ and $\angle C$ at its endpoints. Provided that $\angle B + \angle C < 180^\circ$, the rays $BX$ and $CY$ will not be parallel and will intersect at a unique point $A$. This intersection point $A$ forms the triangle $ABC$. By the ASA congruence criterion, a triangle is uniquely determined by two angles and the included side. Therefore, the construction accurately produces the required triangle.

Example

Answer:

Construction of a Triangle given Two Angles and One Side (AAS Criterion)

Constructing a triangle when the measures of two angles and the length of a side that is *not* between them (a non-included side) are known is the fourth basic triangle construction scenario. This is related to the AAS (Angle-Angle-Side) congruence criterion, which states that if two angles and a non-included side of one triangle are congruent to the corresponding two angles and non-included side of another triangle, then the triangles are congruent. Like other criteria, this implies uniqueness for construction.

Relationship to ASA

The AAS case can always be reduced to the ASA case. In any triangle, the sum of the interior angles is $180^\circ$. If you are given two angles, you can always calculate the third angle.

If you are given $\angle A$ and $\angle B$, you can find $\angle C = 180^\circ - (\angle A + \angle B)$. Similarly, if you are given $\angle B$ and $\angle C$, you can find $\angle A$, and if given $\angle A$ and $\angle C$, you can find $\angle B$.

Now, consider the given non-included side. For example, suppose you are given $\angle A$, $\angle B$, and side $a$ (side $BC$, opposite $\angle A$). You can calculate $\angle C$. With side $BC$ (length $a$) and the two angles adjacent to it, $\angle B$ and $\angle C$, you now have the information for an ASA construction.

Thus, the construction for the AAS case involves one preliminary step: calculating the third angle, and then proceeding with the ASA construction using the given side and the two angles adjacent to it (one given, one calculated).

Construction Steps

Given: The measures of two angles, say $\angle A$ and $\angle B$, and the length of a non-included side, say $a$ (side $BC$, which is opposite to $\angle A$). We aim to construct $\triangle ABC$.

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

Goal: Construct $\triangle ABC$ with angles $\angle A, \angle B$ and side $a=BC$.

Steps:

1. Calculate the Third Angle: Use the angle sum property of a triangle to find the measure of the third angle:

   $\angle C = 180^\circ - (\angle A + \angle B)$

   ... (i)

   Ensure that $\angle A + \angle B < 180^\circ$ for $\angle C$ to be positive and a triangle to exist.
2. Draw the Given Side: Use the ruler to draw the given non-included side. In this example, draw the line segment $BC$ with length equal to $a$. This segment $BC$ is the included side for angles $\angle B$ and the calculated $\angle C$.
3. Construct Angles at Endpoints of the Given Side:
   • At vertex $B$, construct the given angle $\angle B$. Draw a ray $BX$ starting from $B$ such that $\angle CBX = \angle B$.
   • At vertex $C$, construct the calculated angle $\angle C$ (from Step 1). Draw a ray $CY$ starting from $C$ such that $\angle BCY = \angle C$. Ensure $CY$ is drawn on the same side of $BC$ as ray $BX$.
   (If, instead, you were given $\angle B$, $\angle C$, and side $c$ (opposite $\angle C$, i.e., $AB$), you would calculate $\angle A = 180^\circ - (\angle B + \angle C)$, draw side $AB=c$, construct $\angle A$ at $A$ and $\angle B$ at $B$.)
4. Locate the Third Vertex: Extend the rays $BX$ and $CY$ until they intersect. Label the point of intersection as $A$.

The resulting figure is $\triangle ABC$. By construction, side $BC$ has length $a$, $\angle ABC$ is equal to the given $\angle B$, and $\angle ACB$ is equal to the calculated $\angle C$. Since $\angle A + \angle B + \angle C = 180^\circ$, the angle at vertex $A$ is automatically the given $\angle A$. Thus, this is the required triangle.

Justification:

The construction effectively performs an ASA construction using the given side $a=BC$, the given angle $\angle B$ adjacent to it, and the calculated angle $\angle C$ which is also adjacent to it. Since the sum of the given angles is less than $180^\circ$, the calculated third angle is positive, and the rays $BX$ and $CY$ will intersect at a unique point $A$. By the ASA congruence criterion, this uniquely defines the triangle $ABC$. The third angle at $A$ is automatically $180^\circ - (\angle B + \angle C) = 180^\circ - (\angle B + (180^\circ - \angle A - \angle B)) = \angle A$, matching the original given information.

Example

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