Book 3
32
But the angle ABF is also right; therefore the angle ABF is equal to the angles BAD, ABD.
32
Let the angle ABD be subtracted from each; therefore the angle DBF which remains is equal to the angle BAD in the alternate segment of the circle.
32
Next, since ABCD is a quadrilateral in a circle, its opposite angles are equal to two right angles. [III. 22]
32
But the angles DBF, DBE are also equal to two right angles; therefore the angles DBF, DBE are equal to the angles BAD, BCD, of which the angle BAD was proved equal to the angle DBF; therefore the angle DBE which remains is equal to the angle DCB in the alternate segment DCB of the circle.
32
Therefore etc. Q. E. D.
PROPOSITION 33.
33
On a given straight line to describe a segment of a circle admitting an angle equal to a given rectilineal angle.
33
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.
33
The angle at C is then acute, or right, or obtuse.
33
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.
33
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.
33
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]
33
Therefore the circle described with centre G and distance GA will pass through B also.
33
Let it be drawn, and let it be ABE; let EB be joined.
33
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.]
