\(\Delta DEF\) vuông tại D
\(\Rightarrow EF^2=DE^2+DF^2\left(Pythagore\right)\)
\(=3^2+4^2\)
\(=25\)
\(\Rightarrow EF=5\)
\(\Delta DBC\) có EF // BC
\(\Rightarrow\dfrac{EF}{BC}=\dfrac{DE}{DF}\) (định lý Thales)
\(\Rightarrow BC=\dfrac{EF.DF}{DE}\)
\(=\dfrac{5.4}{3}\)
\(=\dfrac{20}{3}\)
\(\Delta DBC\) có EF // BC
\(\Rightarrow\dfrac{EF}{BC}=\dfrac{DF}{DC}\) (hệ quả định lý Thales)
\(\Rightarrow DC=\dfrac{DF.BC}{EF}\)
\(=\dfrac{4.\dfrac{20}{3}}{5}\)
\(=\dfrac{16}{3}\)
\(\Rightarrow x=DC-DF=\dfrac{16}{3}-4=\dfrac{4}{3}\)
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