a,AB = \(\frac{4}{3}DE\Leftrightarrow\frac{AB}{ED}=\frac{4}{3}\)
\(DF=0,75.BC\Leftrightarrow\frac{BC}{DF}=\frac{1}{0,75}=\frac{4}{3}\)
\(\Rightarrow\frac{AB}{ED}=\frac{BC}{DF}=\frac{4}{3}\)
xét \(\Delta ABCvà\Delta EDFtacó\)
\(\frac{AB}{ED}=\frac{BC}{DF}\left(CMT\right)\)
\(\widehat{B}=\widehat{D}\left(gt\right)\)
\(\Rightarrow\Delta ABC\sim\Delta DEF\left(cgc\right)\)
b, theo câu a ta có
\(\frac{AB}{ED}=\frac{BC}{DF}=\frac{AC}{EF}=\frac{4}{3}\)
theo bài ra ta lại có
\(\frac{\Rightarrow AC}{EF}=\frac{EF+5}{EF}=\frac{4}{3}\Rightarrow3EF+15=4EF\)
\(\Rightarrow EF=15cm\)
\(\Rightarrow AC=EF+5=15+5=20cm\)