∣aˉ+bˉ∣−∣aˉ−bˉ∣∣aˉ+bˉ∣+∣aˉ−bˉ∣=2+1
Apply componendo and dividendo
$\begin{aligned}
& \Rightarrow \frac{2|\overline{\mathrm{a}}+\overline{\mathrm{b}}|}{2|\overline{\mathrm{a}}-\overline{\mathrm{b}}|}=\frac{\sqrt{2}+2}{\sqrt{2}} \
& \Rightarrow|\overline{\mathrm{a}}+\overline{\mathrm{b}}|=(1+\sqrt{2})|\overline{\mathrm{a}}-\overline{\mathrm{b}}| \
& \Rightarrow|\overline{\mathrm{a}}+\overline{\mathrm{b}}|^2=(3+2 \sqrt{2})|\overline{\mathrm{a}}-\overline{\mathrm{b}}|^2 \
& \Rightarrow 2|\overline{\mathrm{a}}|^2+2 \overline{\mathrm{a}} \cdot \overline{\mathrm{b}}=(3+2 \sqrt{2})\left(2|\overline{\mathrm{a}}|^2-2 \overline{\mathrm{a}} \cdot \overline{\mathrm{b}}\right) \
& \Rightarrow 2|\overline{\mathrm{a}}|^2(2+2 \sqrt{2})=2 \overline{\mathrm{a}} \cdot \overline{\mathrm{b}}(4+2 \sqrt{2}) \
& \Rightarrow \frac{\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}}{|\overline{\mathrm{a}}|^2}=\frac{2+2 \sqrt{2}}{4+2 \sqrt{2}}=\frac{1}{\sqrt{2}}
\end{aligned}$
Now
$\begin{aligned}
& \frac{|\overline{\mathrm{a}}+\overline{\mathrm{b}}|^2}{|\overline{\mathrm{a}}|^2}=1+\frac{|\overline{\mathrm{b}}|^2}{|\overline{\mathrm{a}}|^2}+\frac{2 \overline{\mathrm{a}} \cdot \overline{\mathrm{~b}}}{|\overline{\mathrm{a}}|^2} \
& =1+1+2\left(\frac{1}{\sqrt{2}}\right)=2+\sqrt{2}
\end{aligned}$