Page:The Elements of Euclid for the Use of Schools and Colleges - 1872.djvu/39

 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; join DE, and on DE, on the side remote from A, describe the equilateral triangle DEF. Join AF. The straight line AF shall bisect the angle BAC. Because AD is equal to AE, 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; therefore the angle DAF is equal to the angle EAF. Wherefore the given rectilineal angle BAC is bisected by the straight line AF. 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, and bisect the angle ACB by the straight line CD, meeting AB at D. AB shall be cut into two equal parts at the point D. Because AC is equal to CB, 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; therefore the base AD is equal to the base DB. Wherefore the given straight line AB is divided into two equal parts at the point D.