Xét ΔBED có AC//ED
nên \(\dfrac{AC}{ED}=\dfrac{BC}{BD}=\dfrac{BA}{BE}\)
=>\(\dfrac{2}{6}=\dfrac{BC}{BC+3,5}=\dfrac{3}{BE}\)
=>\(\left\{{}\begin{matrix}\dfrac{BC}{BC+3,5}=\dfrac{1}{3}\\\dfrac{3}{BE}=\dfrac{1}{3}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}BE=3\cdot3=9\left(cm\right)\\3BC=BC+3,5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}BE=9cm\\2BC=3,5\end{matrix}\right.\)
=>BC=3,5:2=1,75(cm)