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 1/√2 of that side. If then, we take our base, _BC_, in all cases of a length = 2, the transverse partition drawn parallel to it will always have a length equal to 2/√2, or = √2. The vertical {353} partition, _AD_, since _BD_ = 1, will always equal tan β (β being the angle _ABC_). And the oblique partition, _GH_, being equal to _AB_/√2 = 1/(√2 cos β). If then we call our vertical, transverse [Illustration: Fig. 140.] and oblique partitions, _V_, _T_, and _O_, we have _V_ = tan β; _T_ = √2; and _O_ = 1/(√2 cos β), or _V_ : _T_ : _O_ = tan β/√2 : 1 : 1/(2 cos β). And, working out these equations for various values of β, we very soon see that the vertical partition (_V_) is the least of the three until β = 45°, at which limit _V_ and _O_ are each equal to 1/√2 = ·707; and that again, when β = 60°, _O_ and _T_ are each = 1, after which _T_ (whose value always = 1) is the shortest of the three partitions. And, as we have seen, these results are at once applicable, not only to the case of the plane triangle, but also to that of the conical cell. [Illustration: Fig. 141.]