Let BAC be the given rectilineal angle: it is required to bisect it.
Take any point D in AB, and from AC cut off AE equal to AD; [I. 3.
join DE, and on DE, on the side remote from A, describe the equilateral triangle DEF. [I. 1.
Join AF. The straight line AF shall bisect the angle BAC.
Because AD is equal to AE, [Construction.
and AF is common to the two triangles DAF, EAF,
the two sides DA, AF are equal to the two sides EA, AF, each to each;
and the base DF is equal to the base EF; [Definition 24.
therefore the angle DAF is equal to the angle EAF. [I. 8.
Wherefore the given rectilineal angle BAC is bisected by the straight line AF. q.e.f.
PROPOSITION 10. PROBLEM.
To bisect a given finite straight line, that is to divide it into two equal parts.
Let AB be the given straight line; it is required to divide it into two equal parts.
Describe on it an equilateral triangle ABC, [I. 1.
and bisect the angle ACB by the straight line CD, meeting AB at D. [I. 9.
AB shall be cut into two equal parts at the point D.
Because AC is equal to CB, [Definition 24.
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; [Constr.
therefore the base AD is equal to the base DB. [I. 4.
Wherefore the given straight line AB is divided into two equal parts at the point D. q.e.f.
