how to solve part (c)?
You have 2 similar triangles: DEA and BCA.
You know that $\displaystyle DE \parallel BC$
Then the segments of sides must correspond to the lengths of the parallels:
The side AB correspond to BC
The side AD correspond to DE
$\displaystyle \dfrac{AB}{BC} = \dfrac{AD}{DE}$
Since AB = z + k and the measures of the parallels are known to you this proportion becomes:
$\displaystyle \dfrac{z+k}{10} = \dfrac{k}{6}$
Multiply both sides by 60 to get rid of the denominators:
$\displaystyle 6(z+k)=10k~\implies~6z = 4k~\implies~z=\dfrac23k$