§1. In a triangle ABC (fig. 37), BE is made equal to CF; to find the locus of the middle point of EF.

Take K the middle point of BC and P the middle point of EF, then PK is the locus required. For if E′ and F′ are the middle points of BE and CF, the middle point of E′F′ will lie in PK (namely, at the middle point of PK); and again if BE′ and CF′ are bisected in E″ and F″, the middle point of E″F″ will lie in PK (namely, at the middle point of KR); and so on. At any stage we may double the parts cut off from BA and CA instead of bisecting them. Hence the locus required is such that any part of it, however small, contains an infinite number of collinear points; and hence the locus is a straight line.