cate5bf

Let \(BD=x\). By angle bisector theorem, \(\frac{AB}{BD} = \frac{AC}{CD}\) which is equivalent to \(\frac{4}{x}=\frac{5}{3-x} \Rightarrow 12-4x=5x \Rightarrow 9x=12 \Rightarrow x=\frac{4}{3}\). The area of \([\triangle{ADC}]=\frac{DC\cdot AB}{2}=\frac{(3-\frac{4}{3})\cdot4}{2}=\frac{\frac{5}{3}\cdot4}{2}=\frac{10}{3}\approx3\)

\(\triangle{AEM}\) and \(\triangle{B EM}\) share the same base and the same height, so \([\triangle{AEM}]\) = \([\triangle{BEM}]\) and therefore,  \([\triangle{BEM}]\) = \(\frac{1}{2}[\triangle{BAM}]\). \([\triangle{BAM}]\) and \([\triangle{CAM}]\) also share the same base and the same height, so \([\triangle{BAM}]\) = \([\triangle{CAM}]\) and therefore,  \([\triangle{BAM}]\) = \(\frac{1}{2}\triangle{ABC}\). Draw line \(EC\).  \(\triangle{BEM}\) and \(\triangle{CEM}\) also share the same base and the same height, so \([\triangle{BEM}]\) = \([\triangle{CEM}]\) and therefore,  \([\triangle{BEM}]\) = \(\frac{1}{2}[\triangle{BEC}]\). In the end, you get \([\triangle{BEC}]=2[\triangle{CEM}]=2\cdot\frac{1}{2}[\triangle{BAM}]=2\cdot\frac{1}{2}\cdot\frac{1}{2}[\triangle{ABC}]=\frac{1}{2}[\triangle{ABC}]=\frac{1}{2}\cdot14=7\).