In $\mathrm{â–³}CDB$,

${BC}^2={CD}^2+{BD}^2\text{Â}\text{[Pythagoras theorem]}$

${BC}^2={CD}^2+{(DA+AB)}^2$ 

${BC}^2={CD}^2+{DA}^2+{AB}^2+(2×DA×AB)\text{Â}...\left(i\right)$

In $\mathrm{â–³}ADC$,

${CD}^2+{DA}^2={AC}^2\text{Â}...\left(ii\right)\text{Â}\text{[Pythagoras Theorem]}$
Here, $\mathrm{âˆ}CAB={120}^{∘}$ (given)
$⇒\mathrm{âˆ}CAD={60}^{∘}$ (since  $\mathrm{âˆ}CAD$ and $\mathrm{âˆ}CAB$ form a linear pair of angles)

Also, $\cos {60}^{∘}=\frac{AD}{AC}$

$AC=2AD\text{Â}...\left(iii\right)$

Substituting  the values from $\left(ii\right)\text{Â}\&\text{Â}\left(iii\right)\text{Â}\text{in}\text{Â}\left(i\right)$ we get,

${BC}^2={AC}^2+{AB}^2+(AC×AB)$

$a^2=b^2+c^2+bc$

Alternatively,

 Since  $\mathrm{âˆ}A$ is an obtuse angle in  $\mathrm{â–³}ABC$ so,

${BC}^2={AB}^2+{AC}^2+2AB.AD\text{Â}\text{Â}\text{Â}\text{Â}\text{Â}\text{Â}\text{Â}\text{Â}={AB}^2+{AC}^2+2×AB×\frac{1}{2}×AC\text{Â}\text{Â}\text{Â}\text{Â}\text{Â}\text{Â}[âˆ\mu AD=AC\cos {60}^{∘}=\frac{1}{2}AC]\text{Â}\text{Â}\text{Â}\text{Â}\text{Â}\text{Â}\text{Â}\text{Â}={AB}^2+{AC}^2+AB×AC$
 

  $⇒a^2=b^2+c^2+bc$