SOLUTION:
[tex]\textsf{Connect point A and point E.}[/tex]
[tex]\textsf{Observe that AE is the height of } \sf \triangle ABC \: and \: \triangle ABE[/tex]
[tex]\sf Area_{\triangle ABE} = \frac{1}{2}CE \times AE \rightarrow First \: eq.[/tex]
[tex]\sf Area_{\triangle ABC} = \frac{1}{2}(CE + BE) AE[/tex]
[tex]\textsf{Since CE = 2EB,}[/tex]
[tex]\sf Area_{\triangle ABC} = \frac{1}{2}(2BE + BE) AE = \frac{1}{2}(3BE ) AE \rightarrow Second \: eq.[/tex]
[tex]\textsf{Dividing first eq. and second eq. yields}[/tex]
[tex]\sf Area_{\triangle ABE} = \frac{1}{3}Area_{\triangle ABC} = \frac{10}{3} \: cm^2[/tex]
[tex]\textsf{Now, observe that } \sf \triangle ADF \: and \: \triangle ABE \: have \: the \: same \: base[/tex]
[tex]\textsf{So,}[/tex]
[tex]\sf Area_{\triangle ADF} = \frac{1}{2}BE \times BD[/tex]
[tex]\sf Area_{\triangle ABE} = \frac{1}{2} BE(BD+DA)[/tex]
[tex]\textsf{We have BD = 2DA, thus}[/tex]
[tex]\sf Area_{\triangle ABE} = \frac{1}{2} BE(BD+\frac{1}{2}BE)[/tex]
[tex]\textsf{Dividing the areas of the two triangles yields}[/tex]
[tex]\sf Area_{\triangle ADF} = \frac{2}{3}Area_{\triangle ABE}[/tex]
[tex]\sf Area_{\triangle ADF} = \frac{2}{3} \times \frac{10}{3} \: cm^2 = \frac{20}{9} \: cm^2[/tex]
[tex]\textsf{Similarly, we also have AF = 2FC, then AF = BD and FC = DA}[/tex]
[tex]\textsf{This implies that:}[/tex]
[tex]\sf Area_{\triangle BDE} = Area_{\triangle ADF} = Area_{\triangle CEF} = \frac{20}{9} \: cm^2[/tex]
[tex]\textsf{It follows that:}[/tex]
[tex]\sf Area_{\triangle DE\textsf{F}} =Area_{\triangle ABC} - Area_{\triangle BDE} - Area_{\triangle ADF} - Area_{\triangle CEF}[/tex]
[tex]\textsf{Substituting the value of the areas,}[/tex]
[tex]\sf Area_{\triangle DE\textsf{F}} = 10 - \frac{20}{9} - \frac{20}{9} - \frac{20}{9} = \boxed{\sf \frac{10}{3} \: cm^2}[/tex]
ANSWER:
10/3 cm²
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