in point, for here the inôÙvesôÙtiôÙgaôÙtion of its modes of symmetrical subdivision is completely met by an examination of the isosceles triangle which constitutes its plane of symmetry.
The bisection of an isosceles triangle by a line which shall be the shortest possible is a very easy problem. Let ABC be such a triangle of which A is the apex; it may be shewn that, for its shortest line of bisection, we are limited to three cases: viz. to a vertical line AD, bisecting the angle at A and the side BC; to a transverse line parallel to the base BC; or to an oblique line parallel to AB or to AC. The respective magnitudes, or lengths, of these partition lines follow at once from the magnitudes of the angles of our triangle. For we know, to begin with, since the areas of similar figures vary as the squares of their linear dimensions, that, in order to bisect the area, a line parallel to one side of our triangle must always have a length equal to 1ã₤ãã₤ãÿ£¢2 of that side. If then, we take our base, BC, in all cases of a length =ã₤2, the transverse partition drawn parallel to it will always have a length equal to 2ã₤ãã₤ãÿ£¢2, or =ã₤ãÿ£¢2. The vertical {353} partition, AD, since BD =ã₤1, will always equal tanã₤öý (öý being the angle ABC). And the oblique partition, GH, being equal to ABã₤ãã₤ãÿ£¢2 =ã₤1ã₤ãã₤(ãÿ£¢2 cosã₤öý). If then we call our vertical, transverse
and oblique partitions, V, T, and O, we have V =ã₤tanã₤öý; T =ã₤ãÿ£¢2; and O =ã₤1ã₤ãã₤(ãÿ£¢2 cosã₤öý), or
And, working out these equations for various values of öý, we very soon see that the vertical partition (V) is the least of the three until öý =ã₤45ô¯, at which limit V and O are each equal to