PQ is || AB, what does ∠PXA equal?

Then why must ∠BAX = ∠XAP?

Similarly, what about ∠QBX and ∠XBA?

Construction. Now reverse the process. What may we do to ⦞A and B in order to fix X? Then how shall PQ be drawn? Now give the proof.

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.