D, E and F are the points on sides BC,

Solution:

It is given that $A B=5 \mathrm{~cm}, B C=8 \mathrm{~cm}$ and $C A=4 \mathrm{~cm}$.

We have to find $A F, C E$ and $B D$.

Since $A D$ is bisector of $\angle A$

So $\frac{A B}{A C}=\frac{B D}{C D}$

Then,

54=BDBC-BD⇒54=BD8-BD⇒40-5BD=4BD⇒9BD=40

So, $B D=\frac{40}{9}$

Since $B E$ is the bisector of $\angle B$.

So,

ABBC=AEEC⇒ABBC=AC-ECEC

$\frac{5}{8}=\frac{4-C E}{C E}$

$5 C E=32-8 C E$

$5 C E+8 C E=32$

$13 C E=32$

So

$C E=\frac{32}{13} \mathrm{~cm}$

Now since $C F$ is the bisector of $\angle C$

So $\frac{B C}{C A}=\frac{B F}{A F}$

$\frac{8}{4}=\frac{A B-A F}{A F}$

$\frac{8}{4}=\frac{5-A F}{A F}$

$8 A F=20-4 A F$

$12 A F=20$

So 

$3 A F=5 \mathrm{~cm}$

$A F=\frac{5}{3} \mathrm{~cm}$

Hence $A F=\frac{5}{3} \mathrm{~cm}$

$C E=\frac{32}{13} \mathrm{~cm}$

And $B D=\frac{40}{9} \mathrm{~cm}$