Find areas of two triangles in a large triangle Can anyone help me with this problem? I can find area of $\triangle BDE$ easily, but not $\triangle EFA$.

Problem: As shown in the figure, the area of $\triangle ABC$ is $5$, $AE=ED$, and $BD =\frac{2}{3}BC$. Find the areas of triangles $\triangle BDE$ and $\triangle EFA$.


Note. $\triangle ABD$ is a right triangle, but not $\triangle ABF$.

Denote $(MNP)$ the area of $\triangle MNP$. Thus we have:
$(EFA) = \dfrac{(DFA)}{2}= \dfrac{(DCA)-(DCF)}{2}$. And $(DCA) = \dfrac{(ABC)}{3} = \dfrac{5}{3}$. Also, $(BDE) = \dfrac{(BDA)}{2}= \dfrac{1}{2}\cdot \dfrac{2(ABC)}{3}= \dfrac{5}{3}$. Plus: $\dfrac{(DCF)}{(BDF)} = \dfrac{DC}{BD} = \dfrac{1}{2}\Rightarrow \dfrac{1}{2} = \dfrac{(DCF)}{(BDE)+(DEF)}= \dfrac{(DCF)}{\dfrac{5}{3} + \dfrac{(AFD)}{2}}= \dfrac{(DCF)}{\dfrac{5}{3}+ \dfrac{(ADC)-(DCF)}{2}}= \dfrac{(DCF)}{\dfrac{5}{3} + \dfrac{\dfrac{5}{3} - (DCF)}{2}}\Rightarrow (DCF) = 1$ from this equation. Thus: $(BDE) + (EFA) = \dfrac{5}{3} + \dfrac{\dfrac{5}{3}-1}{2} = 2$.
