Constructing and Bisecting Angles

Construction of Angles of Specific Measures (e.g., 60°, 90°, 120°)

Geometric construction allows us to create fundamental angles accurately using only a compass and a straightedge (an unmarked ruler). These basic angles serve as building blocks for constructing many other angles.

Construction of a 60° Angle

The construction of a 60° angle is based on the properties of an equilateral triangle, where all angles are $60^\circ$.

Tools: Compass, Straightedge.

Steps:

1. Draw a Ray: Draw a ray $OA$ starting from point $O$. This will be one arm of the angle.
2. Draw Initial Arc: With $O$ as the center and using any convenient compass radius, draw an arc that intersects the ray $OA$. Let the point of intersection be $P$.
3. Draw Second Arc: Now, with $P$ as the center and keeping the same radius that was used in Step 2, draw another arc that intersects the first arc (the one drawn from $O$). Let the point of intersection of the two arcs be $Q$.
4. Draw the Second Ray: Draw a ray $OB$ starting from $O$ and passing through the point $Q$.

The angle $\angle AOB$ is the required $60^\circ$ angle.

Justification:

By construction, $OP = OQ$ (radii of the arc from $O$) and $OP = PQ$ (arcs drawn with the same radius). Thus, $OP = OQ = PQ$. Triangle $OPQ$ is therefore an equilateral triangle. In an equilateral triangle, all interior angles are equal to $60^\circ$. Hence, $\angle POQ = 60^\circ$. Since $P$ lies on ray $OA$ and $Q$ lies on ray $OB$, the angle $\angle AOB$ is the same as $\angle POQ$. Thus, $\angle AOB = 60^\circ$.

Construction of a 90° Angle (Perpendicular at a Point on a Line)

Constructing a $90^\circ$ angle at a specific point on a line is equivalent to constructing a perpendicular to the line at that point. This construction uses the property of a perpendicular bisector.

Tools: Compass, Straightedge.

Steps:

1. Draw a Line and Point: Draw a line $l$ and mark a point $P$ on it. This is the point where the perpendicular will be constructed.
2. Draw Semicircle/Arc: With $P$ as the center and using any convenient radius, draw an arc (or a semicircle) that intersects the line $l$ at two distinct points. Let these points be $A$ and $B$. Note that $A$ and $B$ are equidistant from $P$, and $P$ is the midpoint of the segment $AB$.
3. Draw Intersecting Arcs (Above/Below Line): With $A$ as the center and using a compass radius greater than the distance $AP$ (or $BP$), draw an arc on one side of the line $l$ (conventionally, above the line).
4. Draw Second Intersecting Arc: With $B$ as the center and using the same radius as in Step 3, draw another arc that intersects the arc drawn from $A$. Let the point of intersection of these two arcs be $Q$.
5. Draw the Perpendicular Ray: Draw the ray $PQ$ starting from $P$ and passing through $Q$.

The ray $PQ$ is perpendicular to the line $l$ at point $P$. The angle formed between $PQ$ and $l$ (e.g., $\angle QPA$ or $\angle QPB$) is $90^\circ$.

Justification:

Consider points $A$, $B$, $P$, and $Q$. By construction, $P$ is the midpoint of the segment $AB$ ($AP = BP$). Also by construction, $QA = QB$ (arcs drawn with the same radius from $A$ and $B$). A point equidistant from the endpoints of a line segment lies on the perpendicular bisector of the segment. Since both $P$ and $Q$ are equidistant from $A$ and $B$, the line $PQ$ is the perpendicular bisector of the segment $AB$. As $AB$ lies on line $l$, the line $PQ$ is perpendicular to line $l$ at $P$. Therefore, the angle $\angle QPA = 90^\circ$.

Alternative Method (using 60° and 120°):

A $90^\circ$ angle can also be constructed by first constructing a $60^\circ$ angle and a $120^\circ$ angle (see the next construction) sharing the same vertex and initial arm. The region between the ray marking $60^\circ$ and the ray marking $120^\circ$ corresponds to an angle of $120^\circ - 60^\circ = 60^\circ$. Bisecting this $60^\circ$ angle (using the angle bisection method described in I3) adds $30^\circ$ to the $60^\circ$ angle, resulting in an angle of $60^\circ + 30^\circ = 90^\circ$.

Construction of a 120° Angle

This construction is an extension of the $60^\circ$ angle construction.

Tools: Compass, Straightedge.

Steps:

1. Draw a Ray: Draw a ray $OA$ starting from point $O$. This will be the initial arm.
2. Draw Initial Arc: With $O$ as the center and any convenient compass radius, draw an arc that intersects the ray $OA$ at a point $P$.
3. Mark 60°: With $P$ as the center and using the same radius as in Step 2, draw an arc intersecting the first arc (from $O$) at a point $Q$. The angle $\angle AOQ$ is $60^\circ$.
4. Mark 120°: Now, with $Q$ as the center and keeping the same radius, draw another arc that intersects the first arc (from $O$) at a point $R$. The angle $\angle QOR$ is also $60^\circ$.
5. Draw the Second Ray: Draw the ray $OC$ starting from $O$ and passing through the point $R$.

The angle $\angle AOC$ is the required $120^\circ$ angle.

Justification:

By construction, $\triangle OPQ$ is equilateral, so $\angle POQ = 60^\circ$. Similarly, $\triangle OQR$ is constructed with $OQ = QR = RO$ (all equal to the common radius), so $\triangle OQR$ is also equilateral, and $\angle QOR = 60^\circ$. The angle $\angle AOC$ is the sum of the adjacent angles $\angle AOQ$ and $\angle QOR$.

Thus, the angle $\angle AOC$ measures $120^\circ$.

Examples

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Construction of Angles using Compass and Ruler (derived angles like 30°, 45°, 75°, 105°, 150°)

Many other specific angles can be constructed accurately using compass and straightedge by combining the basic constructions ($60^\circ$, $90^\circ$, $120^\circ$) with a powerful technique called angle bisection. Any angle that can be expressed as $m \times 60^\circ + n \times 90^\circ$ divided by $2^k$ (where m, n, k are non-negative integers) can theoretically be constructed.

Angle Bisection

The key technique for constructing many derived angles is angle bisection. An angle bisector divides an angle into two equal halves. The method for constructing an angle bisector is detailed in the next section (I3).

Derived Angle Constructions

Here are some common derived angles and how they can be constructed by combining basic angles and angle bisection:

• $30^\circ$ Angle: Construct a $60^\circ$ angle. Then, bisect the $60^\circ$ angle. Each of the resulting angles will be $30^\circ$ ($30^\circ = 60^\circ / 2$).
• $45^\circ$ Angle: Construct a $90^\circ$ angle. Then, bisect the $90^\circ$ angle. Each of the resulting angles will be $45^\circ$ ($45^\circ = 90^\circ / 2$).
• $15^\circ$ Angle: Construct a $30^\circ$ angle (by bisecting $60^\circ$). Then, bisect the $30^\circ$ angle. Each of the resulting angles will be $15^\circ$ ($15^\circ = 30^\circ / 2 = 60^\circ / 4$).
• $75^\circ$ Angle: Construct a $90^\circ$ angle and a $60^\circ$ angle sharing the same vertex and the same initial ray. The angle between the rays forming $60^\circ$ and $90^\circ$ is $90^\circ - 60^\circ = 30^\circ$. Bisect this $30^\circ$ angle. The ray that bisects this $30^\circ$ angle will be at $60^\circ + 15^\circ = 75^\circ$ from the initial ray. ($75^\circ = 60^\circ + (90^\circ - 60^\circ)/2$).
• $105^\circ$ Angle: Construct a $90^\circ$ angle and a $120^\circ$ angle sharing the same vertex and the same initial ray. The angle between the rays forming $90^\circ$ and $120^\circ$ is $120^\circ - 90^\circ = 30^\circ$. Bisect this $30^\circ$ angle. The ray that bisects this $30^\circ$ angle will be at $90^\circ + 15^\circ = 105^\circ$ from the initial ray. ($105^\circ = 90^\circ + (120^\circ - 90^\circ)/2$).
• $135^\circ$ Angle: Construct a $90^\circ$ angle. A straight line forms an angle of $180^\circ$. Consider the line. Construct a $90^\circ$ angle. The angle between the $90^\circ$ ray and the ray pointing in the opposite direction on the line is $180^\circ - 90^\circ = 90^\circ$. Bisect this $90^\circ$ angle. The resulting ray will be at $90^\circ + 45^\circ = 135^\circ$ from the initial ray. ($135^\circ = 90^\circ + 90^\circ/2$).
• $150^\circ$ Angle: Construct a $120^\circ$ angle. Consider the straight line forming $180^\circ$. The angle between the $120^\circ$ ray and the opposite direction ray is $180^\circ - 120^\circ = 60^\circ$. Bisect this $60^\circ$ angle. The resulting ray will be at $120^\circ + 30^\circ = 150^\circ$ from the initial ray. ($150^\circ = 120^\circ + (180^\circ - 120^\circ)/2 = 120^\circ + 60^\circ/2$).
• $22.5^\circ$ Angle: Construct a $45^\circ$ angle (by bisecting $90^\circ$). Then, bisect the $45^\circ$ angle. Each resulting angle is $22.5^\circ$ ($22.5^\circ = 45^\circ / 2 = 90^\circ / 4$).

By repeatedly applying the bisection process to angles constructed from $60^\circ$ and $90^\circ$, a wide range of specific angles can be accurately constructed.

Examples

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Construction of Angle Bisector

An angle bisector is a ray that originates from the vertex of an angle and divides the angle into two angles of equal measure.

Constructing an angle bisector is a fundamental skill used in constructing many other angles and figures.

Construction Steps

Given: An angle $\angle ABC$, with vertex at $B$ and arms $BA$ and $BC$.

Tools: Compass, Straightedge.

Goal: Construct the ray $BD$ such that it bisects $\angle ABC$, meaning $\angle ABD = \angle DBC$.

Steps:

1. Draw Initial Arc: Place the pointed end of the compass on the vertex $B$. Using any convenient compass radius, draw an arc that intersects both arms of the angle, ray $BA$ and ray $BC$. Let the point where the arc intersects $BA$ be $P$ and the point where it intersects $BC$ be $Q$. (Note: By this step, $BP = BQ$ as they are radii of the same arc).
2. Draw Intersecting Arcs from P and Q: Now, with $P$ as the center and using a compass radius (this radius can be the same as in Step 1 or a different one, but it must be large enough for the arcs to intersect within the angle), draw an arc in the interior of $\angle ABC$.
3. Draw Second Intersecting Arc: Keeping the same compass width that was used in Step 2, place the pointed end of the compass on $Q$. Draw another arc that intersects the arc drawn from $P$ in Step 2. Let the point of intersection of these two arcs be $D$.
4. Draw the Bisector Ray: Use the straightedge to draw a ray $BD$ starting from the vertex $B$ and passing through the intersection point $D$.

The ray $BD$ is the angle bisector of $\angle ABC$.

Example

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Justification of Angle Bisector Construction

The justification for the angle bisector construction proves that the ray $BD$, constructed according to the steps in I3, indeed divides the angle $\angle ABC$ into two equal angles, $\angle ABD$ and $\angle DBC$. This is typically done using the concept of congruent triangles.

Proof using Congruent Triangles

Given: Angle $\angle ABC$. Ray $BD$ constructed as per the steps in I3. $P$ is a point on $BA$ and $Q$ is a point on $BC$ such that $BP = BQ$ (from Step 1 of construction). $D$ is a point in the interior of $\angle ABC$ such that $PD = QD$ (from Steps 2 and 3 of construction).

To Prove: Ray $BD$ bisects $\angle ABC$, i.e., $\angle ABD = \angle DBC$. This is equivalent to proving $\angle PBD = \angle QBD$ since $P$ is on $BA$ and $Q$ is on $BC$.

Construction for Proof: Join points $P$ and $Q$ to the point $D$ to form line segments $PD$ and $QD$. This creates two triangles, $\triangle BPD$ and $\triangle BQD$.

Proof:

Statement	Reason
In $\triangle BPD$ and $\triangle BQD$:
$BP = BQ$	Radii of the same arc drawn from center $B$ (Construction Step 1).
$PD = QD$	Arcs of equal radii drawn from $P$ and $Q$ (Construction Steps 2 & 3).
$BD = BD$	Common side to both triangles.
$\triangle BPD \cong \triangle BQD$	By SSS (Side-Side-Side) Congruence rule.
$\angle PBD = \angle QBD$	Corresponding Parts of Congruent Triangles (CPCT).

Since $\angle PBD$ is the same as $\angle ABD$ (as $P$ lies on ray $BA$) and $\angle QBD$ is the same as $\angle DBC$ (as $Q$ lies on ray $BC$), the congruence $\angle PBD = \angle QBD$ implies $\angle ABD = \angle DBC$.

Therefore, the ray $BD$ divides the angle $\angle ABC$ into two equal angles, which means $BD$ is the angle bisector of $\angle ABC$.