Ta có: $\frac{{{a}^{2}}+{{b}^{2}}}{{{b}^{2}}+{{c}^{2}}}=\frac{a}{c}$
$\left( {{a}^{2}}+{{b}^{2}} \right).c=\left( {{b}^{2}}+{{c}^{2}} \right).a$
${{a}^{2}}c+{{b}^{2}}c={{b}^{2}}a+{{c}^{2}}a$
${{a}^{2}}c+{{b}^{2}}c-{{b}^{2}}a-{{c}^{2}}a=0$
$\left( {{a}^{2}}c-{{c}^{2}}a \right)+\left( {{b}^{2}}c-{{b}^{2}}a \right)=0$
$ac\left( a-c \right)-{{b}^{2}}\left( a-c \right)=0$
$\left( a-c \right)\left( ac-{{b}^{2}} \right)=0$
$ac-{{b}^{2}}=0$ (vì $a\ne c$)
$ac={{b}^{2}}$
Mặt khác $\frac{\overline{ab}}{\overline{bc}}=\frac{b}{c}$
$\frac{10a+b}{10b+c}=\frac{b}{c}$
$c\left( 10a+b \right)=b\left( 10b+c \right)$
$10ac+bc=10{{b}^{2}}+bc$
$ac={{b}^{2}}$ (luôn đúng)
Vậy $\frac{\overline{ab}}{\overline{bc}}=\frac{b}{c}$