$a)$

$tw.Pitagorasadla\Delta ADC$

${\left|AD\right|}^2+8^2={10}^2$

${\left|AD\right|}^2+64=100\mid -64$

${\left|AD\right|}^2=36$

$\left|AD\right|=\sqrt{36}=6cm$

$tw.Pitagorasadla\Delta ABC$

${\left|AC\right|}^2+{\left|AB\right|}^2={\left|BC\right|}^2$

$8^2+{\left(6+9\right)}^2={\left|BC\right|}^2$

$64+225={\left|BC\right|}^2$

${\left|BC\right|}^2=289$

$\left|BC\right|=\sqrt{289}=17cm$

$\left|BC\right|-\left|CD\right|=17cm-10cm=7cm$

$b)$

$tw.Pitagorasadla\Delta ABC$

${12}^2+{16}^2={\left|BC\right|}^2$

$144+256={\left|BC\right|}^2$

${\left|BC\right|}^2=400$

$\left|BC\right|=\sqrt{400}=20cm$

$\left|\angle BAC\right|=\left|\angle DEC\right|,\left|\angle ABC\right|=\left|\angle DEC\right|,\left|\angle ACB\right|=\left|\angle DCE\right|\Rightarrow \Delta ABC\sim \Delta EDC\left(cechakkk\right)$

$\frac{\left|BC\right|}{\left|CD\right|}=\frac{\left|AC\right|}{\left|CE\right|}=\frac{\left|BA\right|}{\left|ED\right|}$

$\frac{20}{\left|CD\right|}=\frac{16}{\left|CE\right|}=\frac{12}{9}$

$\frac{20}{\left|CD\right|}=\frac{12}{9}\Rightarrow 9\cdot 20=12\left|CD\right|\Rightarrow \left|CD\right|=\frac{9\cdot 20}{12}=\frac{3\cdot 20}{4}=3\cdot 5=15cm$

$\left|BC\right|-\left|CD\right|=20cm-15cm=5cm$