Question

Va rog sa ma ajutati. Problema este din triunghiul asemenea
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Answer 1

FE = FA + EA = 2 + 8 = 10

AB || DE ⇒ ΔFAB ~ ΔFED

$\dfrac{FA}{FE} = \dfrac{BA}{DE} = \dfrac{FB}{FD} \iff \dfrac{2}{10} = \dfrac{5}{DE} = \dfrac{4}{FD}$

$\dfrac{2}{10} = \dfrac{1}{5}$

$\dfrac{1}{5} = \dfrac{5}{DE} \implies DE = 25$

$\dfrac{1}{5} = \dfrac{4}{FD} \implies FD = 20$

Answer 2

$\it EF=EA+AF=8+2=10\ cm\\ \\ \\ AB||DF\ \stackrel{T.f.a.}{\Longrightarrow}\ \Delta FDE \sim\Delta FBA \Rightarrow \dfrac{FD}{FB}=\dfrac{DE}{BA}=\dfrac{FE}{FA} \Rightarrow\\ \\ \\ \Rightarrow \dfrac{FD}{4}=\dfrac{DE}{5}=\dfrac{10}{2}=5 \Rightarrow \begin{cases}\it FD=4\cdot5=20\ cm\\ \\ \it DE=5\cdot5=25\ cm \end{cases}$