Let ADB be the given arc: it is required to bisect it.
Join AB;
bisect it at C; [I. 10.
from the point C draw CD at right angles to AB meeting the arc at D. [I. 11
The arc ADB shall be bisected at the point D.
Join AD, DB.
Then, because AC is equal to CB, [Construction.
and CD is common to the two triangles ACD, BCD;
the two sides AC, CD are equal to the two sides BC, CD, each to each;
and the angle ACD is equal to the angle BCD, because each of them is a right angle; [Construction.
therefore the base AD is equal to the base BD. [I. 4.
But equal straight lines cut off equal arcs, the greater equal to the greater, and the less equal to the less; [III. 28.
and each of the arcs AD, DB is less than a semi-circumference, because DC, if produced, is a diameter; [III. 1. Cor. therefore the arc AD is equal to the arc DB.
Wherefore the given arc is bisected at D. q.e.f.
PROPOSITION 31. THEOREM.
In a circle the angle in a semicircle is a right angle; but the angle in a segment greater than a semicircle is less than a right angle; and the angle in a segment less than a semicircle is greater than a right angle.
Let ABCD be a circle, of which BC is a diameter and E the centre; and draw CA, dividing the circle into the segments ABC, ADC, and join BA, AD, DC: the angle in the semicircle BAC shall be a right angle; but the angle in the segment ABC, which is greater than a
