DF and AC are similar
and
BC and EF are similar sides
So make a proportion
[tex]\sf{ \frac{AC}{DF} = \frac{BC}{EF} }[/tex]
Plug in the numbers
[tex]\sf{ \frac{9}{DF} = \frac{15}{25} }[/tex]
Cross multiply
[tex]\sf{DF \times 15 = 9 \times 25}[/tex]
Isolate DF and simplify it
[tex]\sf{DF = 15}[/tex]
And so your final answer is
[tex]\boxed{\bf{15 ~centimeters}}[/tex]
