In $ \Delta ABC , m\angle CBA = 72 . $ $E$ is the midpoint of $AC$ and $D$ is on $BC$ such that $2BD = DC$.$AD \cap BE = {F}$.
Area of $\Delta BDF = 10 . $ Find area of $\Delta DEF$
2 Answers
$\begingroup$Let $K$ be a midpoint of $DC$.
Thus, since $E$ is a midpoint of $AC$, we obtain that $EK||AD$ and since $D$ is a midpoint of $BK$, we obtain $BF=FE$ and $$S_{\Delta FDE}=S_{\Delta BDF}=10.$$
$\endgroup$ $\begingroup$The areas will be the same; $[\Delta DEF]=10$. We begin my using mass points letting $F$ be the center of mass. If we let $B$ have a mass of 2 so that $C$ has a mass of 1, then $A$ must also have a mass of 1 since $E$ is the midpoint. Then $DF/FA=1/3$ so $[\Delta BFA]=30$. Similarly, $BF/FE=1$ since $B$ and $E$ both have mass $2$. Thus, $[\Delta AFE]=30$ as well. Then using $DF/FA=1/3$ again, $[\Delta DFE]=10$
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