The third general method of attack applies chiefly to problems where some point is to be determined. This is the method of the intersection of loci. Thus, to locate an electric light at a point eighteen feet from the point of intersection of two streets and equidistant from them, evidently one locus is a circle with a radius eighteen feet and the center at the vertex of the angle made by the streets, and the other locus is the bisector of the angle. The method is also occasionally applicable to theorems. For example, to prove that the perpendicular bisectors of the sides of a triangle are concurrent. Here the locus of points equidistant from _A_ and _B_ is _PP'_, and the locus of points equidistant from _B_ and _C_ is _QQ'_. These can easily be shown to intersect, as at _O_. Then _O_, being equidistant from _A_, _B_, and _C_, is also on the perpendicular bisector of _AC_. Therefore these bisectors are concurrent in _O_. These are the chief methods of attack, and are all that should be given to an average class for practical use. Besides the methods of attack, there are a few general directions that should be given to pupils. 1. In attacking either a theorem or a problem, take the most general figure possible. Thus, if a proposition relates to a quadrilateral, take one with unequal sides and unequal angles rather than a square or even a rectangle. The simpler figures often deceive a pupil into feeling that he has a proof, when in reality he has one only for a special case. 2. Set forth very exactly the thing that is given, using letters relating to the figure that has been drawn. Then set forth with the same exactness the thing that is to be proved. The neglect to do this is the cause of a large per cent of the failures. The knowing of exactly what we have to do and exactly what we have with which to do it is half the battle. 3. If the proposition seems hazy, the difficulty is probably with the wording. In this case try substituting the definition for the name of the thing defined. Thus instead of thinking too long about proving that the median to the base of an isosceles triangle is perpendicular to the base, draw the figure and think that there is given _AC_ = _BC_, _AD_ = _BD_, and that there is to be proved that [L]_CDA_ = [L]_BDC_. [Illustration] Here we have replaced "median," "isosceles," and "perpendicular" by statements that express the same idea in simpler language.