If $|2z-1|=|z-2|$  and $z_1,z_2$  are complex numbers such that $|z_1-\alpha |<\alpha ,|z_2-\beta |<\beta ,$  then $|z_1+z_2/\alpha +\beta |$ 

$|2z-1|=|z-2|$

$\Rightarrow$  $|2z-1{|}^2=|z-2{|}^2$

$\Rightarrow$  $(2z-1)(2\overline{z}-1)=(z-2)(\overline{z}-2)$

$\Rightarrow$ $4z\overline{z}-2\overline{z}-2z+1=z\overline{z}-2\overline{z}-2z+4$

$\Rightarrow 3|z{|}^2=3$

$\Rightarrow$ $\left|z\right|=1$

Again,

$|z_1+z_2|=|z_1-\alpha +z_2-\beta +\alpha +\beta |$

                $\le |z_1-\alpha |+|z_2-\beta |+|\alpha +\beta |$

                $<\alpha +\beta +|\alpha +\beta |=2|\alpha +\beta |[\because \alpha ,\beta >0]$

$\therefore \left|\frac{z_1+z_2}{\alpha +\beta }\right|<2$

$\Rightarrow \left|\frac{z_1+z_2}{\alpha +\beta }\right|<2\left|z\right|$