Ta có:
\(\begin{array}{l}\dfrac{{OA}}{{OB}} = \dfrac{9}{{12}} = \dfrac{3}{4}\\\dfrac{{OB}}{{OC}} = \dfrac{{12}}{{16}} = \dfrac{3}{4}\\ \Rightarrow \dfrac{{OA}}{{OB}} = \dfrac{{OB}}{{OC}} = \dfrac{3}{4}\end{array}\)
Xét \(\Delta OAB\) và \(\Delta OBC\) có:
\(\dfrac{{OA}}{{OB}} = \dfrac{{OB}}{{OC}} = \dfrac{3}{4}\) (cmt)
\(\widehat {AOB} = \widehat {BOC}\) (vì \(Oz\) là tia phân giác của \(\widehat {xOy}\))
\( \Rightarrow \Delta OAB\) đồng dạng \(\Delta OBC\) (c.g.c).
\(\begin{array}{l} \Rightarrow \dfrac{{AB}}{{BC}} = \dfrac{{OA}}{{OB}} = \dfrac{3}{4}\\ \Rightarrow BC = \dfrac{4}{3}AB = \dfrac{4}{3}.6 = 8\,\left( {cm} \right)\end{array}\)
Chọn B.
