I don't know what to do on this question and would appreciate detailed help.
Thanks in advance!
triangles ADB and BDC are similar.
so,
$\displaystyle \frac{AD}{BD}=\frac{DB}{DC}$
$\displaystyle \frac{AD}{12}=\frac{12}{5}$
$\displaystyle AD=\frac{144}{5}=28.8\;\;cm$
$\displaystyle AC=AD+DC=28.8+5=33.8 \;\;cm$
$\displaystyle ED = AC-(14+5)=33.8-19=14.8\;\;cm$
$\displaystyle EC=AC-AE=33.8=14=19.8\;\;cm$
In triangle EBD,
$\displaystyle EB^2=BD^2+ED^2$
$\displaystyle EB=\sqrt{(12)^2+(14.8)^2}$
EB = 19.05 cm
NOW,
$\displaystyle FE \times EB=AE \times EC$
$\displaystyle FE \times 19.05=14 \times 19.8$
FE = 14.6 cm
got it ???