Construction of Segment on Given Line Admitting Given Angle

Theorem
On any given line segment, it is possible to describe a segment of a circle which admits an angle equal to any given rectilineal angle.

Construction
In the following:
 * let $AB$ be the line segment on which the segment is to be drawn;
 * let $C$ be the rectilineal angle to which the angle in the segment is to be made equal.

There are three cases to consider:
 * $(1): \quad$ When $C$ is an acute angle
 * $(2): \quad$ When $C$ is a right angle
 * $(3): \quad$ When $C$ is an obtuse angle.

Each one will be addressed in turn.

Acute Angle
Construct $\angle BAD$ on $AB$ such that $\angle BAD = \angle C$ using Construction of Equal Angle.

Draw $AK$ perpendicular to $AD$.

Bisect $AD$ at $F$.

Draw $FG$ perpendicular to $AD$ where $G$ is on $AK$.

Join $GB$.

Draw the circle whose center is at $G$ and whose radius is $GB$.

Let this circle meet $AK$ at $E$.

The required segment is the one whose base is $AB$ and whose circumference is the greater of the two arcs between $A$ and $B$.


 * Euclid-III-33a.png

Right Angle
Construct $\angle BAD$ on $AB$ such that $\angle BAD = \angle C$ using Construction of Equal Angle.

Bisect $AB$ at $F$.

Draw the circle whose center is at $F$ and whose radius is $AF$ (or $BF$).

The required segment is the one whose base is $AB$ and whose circumference is the either of the two arcs between $A$ and $B$.


 * Euclid-III-33r.png

Obtuse Angle
Construct $\angle BAD$ on $AB$ such that $\angle BAD = \angle C$ using Construction of Equal Angle.

Draw $AK$ perpendicular to $AD$.

Bisect $AD$ at $F$.

Draw $FG$ perpendicular to $AD$ where $G$ is on $AK$.

Join $GB$.

Draw the circle whose center is at $G$ and whose radius is $GB$.

Let this circle meet $AK$ at $E$.

The required segment is the one whose base is $AB$ and whose circumference is the lesser of the two arcs between $A$ and $B$.


 * Euclid-III-33o.png

Proof for Acute Angle
We have that $AF = FB$, $\angle AFG = \angle BFG$ (both are right angles) and $AG$ is common.

So from Triangle Side-Angle-Side Equality it follows that $AG = BG$.

So the circle $ABE$ whose center is at $G$ and whose radius is $GB$ also passes through $A$.

As $AE$ passes through the center $G$, $AE$ is a diameter of the circle $ABE$.

As $AD$ is perpendicular to $AE$, it follows from Line at Right Angles to Diameter of Circle that $AD$ is tangent to circle $ABE$.

From the Tangent-Chord Theorem, $\angle DAB = \angle AEB$.

But $\angle DAB = \angle C$ by construction.

Hence the result.

Proof for Right Angle
Select a point $E$ on the circle and join $AE$ and $BE$.

As $AD$ is perpendicular to $AB$, it follows from Line at Right Angles to Diameter of Circle that $AD$ is tangent to circle $ABE$.

From Relative Sizes of Angles in Segments, $\angle BAD = \angle AEB$.

But as $\angle BAD = \angle C$ by construction, $\angle AEB = \angle C$.

Hence the result.

Proof for Obtuse Angle
We have that $AF = FB$, $\angle AFG = \angle BFG$ (both are right angles) and $AG$ is common.

So from Triangle Side-Angle-Side Equality it follows that $AG = BG$.

So the circle $ABE$ whose center is at $G$ and whose radius is $GB$ also passes through $A$.

As $AE$ passes through the center $G$, $AE$ is a diameter of the circle $ABE$.

As $AD$ is perpendicular to $AE$, it follows from Line at Right Angles to Diameter of Circle that $AD$ is tangent to circle $ABE$.

Now place point $H$ anywhere on the arc $AB$ and join $AH$ and $HB$.

From the Tangent-Chord Theorem, $\angle DAB = \angle AHB$.

But $\angle DAB = \angle C$ by construction.

Hence the result.