PROPOSITION 33.
On a given straight line to describe a segment of a circle admitting an angle equal to a given rectilineal angle.
Let
AB be the given straight line, and the angle at
C the given rectilineal angle; thus it is required to describe on the given straight line
AB a segment of a circle admitting an angle equal to the angle at
C.
The angle at
C is then acute, or right, or obtuse.
First let it be acute, and, as in the first figure, on the straight line
AB, and at the point
A, let the angle
BAD be constructed equal to the angle at
C;
therefore the angle BAD is also acute.
Let
AE be drawn at right angles to
DA, let
AB be bisected at
F, let
FG be drawn from the point
F at right angles to
AB, and let
GB be joined.
Then, since
AF is equal to
FB, and
FG is common,
the two sides AF, FG are equal to the two sides BF, FG; and the angle
AFG is equal to the angle
BFG;
therefore the base AG is equal to the base BG. [I. 4]
Therefore the circle described with centre
G and distance
GA will pass through
B also.
Let it be drawn, and let it be
ABE; let
EB be joined.
Now, since
AD is drawn from
A, the extremity of the diameter
AE, at right angles to
AE,
therefore AD touches the circle ABE. [III. 16, Por.]
Since then a straight line
AD touches the circle
ABE, and from the point of contact at
A a straight line
AB is drawn across in the circle
ABE,
the angle DAB is equal to the angle AEB in the alternate segment of the circle. [III. 32]
But the angle
DAB is equal to the angle at
C; therefore the angle at
C is also equal to the angle
AEB.
Therefore on the given straight line
AB the segment
AEB of a circle has been described admitting the angle
AEB equal to the given angle, the angle at
C.
Next let the angle at
C be right; and let it be again required to describe on
AB a segment of a circle admitting an angle equal to the right angle at
C.
Let the angle
BAD be constructed equal to the right angle at
C, as is the case in the second figure; let
AB be bisected at
F, and with centre
F and distance either
FA or
FB let the circle
AEB be described.
Therefore the straight line
AD touches the circle
ABE, because the angle at
A is right. [
III. 16, Por.]
And the angle
BAD is equal to the angle in the segment
AEB, for the latter too is itself a right angle, being an angle in a semicircle. [
III. 31]
But the angle
BAD is also equal to the angle at
C.
Therefore the angle
AEB is also equal to the angle at
C.
Therefore again the segment
AEB of a circle has been described on
AB admitting an angle equal to the angle at
C.
Next, let the angle at
C be obtuse; and on the straight line
AB, and at the point
A, let the angle
BAD be constructed equal to it, as is the case in the third figure; let
AE be drawn at right angles to
AD, let
AB be again bisected at
F, let
FG be drawn at right angles to
AB, and let
GB be joined.
Then, since
AF is again equal to
FB, and
FG is common,
the two sides AF, FG are equal to the two sides BF, FG; and the angle
AFG is equal to the angle
BFG;
therefore the base AG is equal to the base BG. [I. 4]
Therefore the circle described with centre
G and distance
GA will pass through
B also; let it so pass, as
AEB.
Now, since
AD is drawn at right angles to the diameter
AE from its extremity,
AD touches the circle AEB. [III. 16, Por.]
And
AB has been drawn across from the point of contact at
A;
therefore the angle BAD is equal to the angle constructed in the alternate segment AHB of the circle. [III. 32]
But the angle
BAD is equal to the angle at
C.
Therefore the angle in the segment
AHB is also equal to the angle at
C.
Therefore on the given straight line
AB the segment
AHB of a circle has been described admitting an angle equal to the angle at
C. Q. E. F.