10mm =1 cm
\(\Delta ABO\infty\Delta A'B'O\Rightarrow\dfrac{AB}{A'B'}=\dfrac{OA}{OA'}\left(1\right)\)\(\Delta OIF\infty\Delta A'B'F\Rightarrow\dfrac{OF}{A'F}=\dfrac{OI}{A'B'}\)
\(\Rightarrow\dfrac{OF}{OF-OA'}=\dfrac{AB}{A'B'}\left(2\right)\)
(1,2) \(\Rightarrow\dfrac{OF'}{OF-OA'}=\dfrac{OA}{OA'}\Rightarrow OA'=6\)
=> A'B' = 0,5 cm