ABC is a triangle such that \(\angle ABC = 120^\circ\). If BD is the bisector of \(\angle B\) that meets AC at D, then what is the ratio of the area of \(\triangle ABD\) to the area of \(\triangle CBD\)?
\(AB : BC\)
Since BD bisects \(\angle ABC\), \(\angle ABD = \angle DBC = 60^\circ\). So area of \(\triangle ABD = \tfrac12\cdot AB\cdot BD\cdot\sin 60^\circ\) and area of \(\triangle CBD = \tfrac12\cdot BC\cdot BD\cdot\sin 60^\circ\). Taking the ratio, the common factors BD and \(\sin 60^\circ\) cancel, giving \(\dfrac{[\triangle ABD]}{[\triangle CBD]} = AB : BC\). (Equivalently, by the angle bisector theorem D divides AC in the ratio AB : BC, and both triangles share the same height from B, so their areas are in the ratio of their bases AD : DC = AB : BC.)
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