Ta có \(S\)ADFE = DF.FE \(\left(1\right)\)

Xét \(\Delta DBF\) và \(\Delta E\)\(FC\) có :

\(\hat{BDF}\) \(=\) \(\hat{FEC}\) \(\left(=90°\right)\)

\(\hat{DFB}\) \(=\) \(\hat{ECF}\) ( cùng phụ với \(\hat{EFC}\))

\(\Rightarrow\Delta DBF\approx\Delta EFC\)(\(G-G\) )

\(\Rightarrow\dfrac{DF}{EC}=\dfrac{BD}{FE}\)

\(\Leftrightarrow\dfrac{DF}{25}=\dfrac{20}{FE}\)

\(\Rightarrow DF.FE=20.25=500\) \(\left(2\right)\)

\(\underrightarrow{\left(1\right)\left(2\right)}\)   \(S\)ADFE =500 (\(cm^2\))