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 $B$.
:
=== 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 $B$.
:
=== 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 $B$.
:
== Proof ==
=== 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 Congruence 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 Congruence 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.
== Sources ==
* : Book $\text{III}$. Propositions
Category:Circles