Book 3
29
Now, since the circumference BGC is equal to the circumference EHF, the angle BKC is also equal to the angle ELF. [III. 27]
29
And, since the circles ABC, DEF are equal, the radii are also equal; therefore the two sides BK, KC are equal to the two sides EL, LF; and they contain equal angles; therefore the base BC is equal to the base EF. [I. 4]
29
Therefore etc.
PROPOSITION 30.
30
To bisect a given circumference.
30
Let ADB be the given circumference; thus it is required to bisect the circumference ADB.
30
Let AB be joined and bisected at C; from the point C let CD be drawn at right angles to the straight line AB, and let AD, DB be joined.
30
Then, since AC is equal to CB, and CD is common, the two sides AC, CD are equal to the two sides BC, CD; and the angle ACD is equal to the angle BCD, for each is right; therefore the base AD is equal to the base DB. [I. 4]
30
But equal straight lines cut off equal circumferences, the greater equal to the greater, and the less to the less; [III. 28] and each of the circumferences AD, DB is less than a semicircle; therefore the circumference AD is equal to the circumference DB.
30
Therefore the given circumference has been bisected at the point D. Q. E. F.
PROPOSITION 31.
31
In a circle the angle in the semicircle is right, that in a greater segment less than a right angle, and that in a less segment greater than a right angle; and further the angle of the greater segment is greater than a right angle, and the angle of the less segment less than a right angle.
