Đặt $\frac{a}{b}=\frac{c}{d}=t\left( t\ne 0 \right)\Rightarrow \left\{ \begin{align}
& a=b.t \\
& c=d.t \\
\end{align} \right.$
a) $\Rightarrow \frac{a}{a+b}=\frac{b.t}{b.t+b}=\frac{b.t}{b\left( t+1 \right)}=\frac{t}{t+1}$
$\Rightarrow \frac{c}{c+d}=\frac{d.t}{d.t+d}=\frac{d.t}{d\left( t+1 \right)}=\frac{t}{t+1}$
$\Rightarrow \frac{a}{a+b}=\frac{t}{t+1}=\frac{c}{c+d}\Rightarrow \frac{a}{a+b}=\frac{c}{c+d}$
b) $\frac{2a+b}{2a-b}=\frac{2.b.t+b}{2.b.t-b}=\frac{b\left( 2t+1 \right)}{b\left( 2t-1 \right)}=\frac{2t+1}{2t-1}$
$\frac{2c+d}{2c-d}=\frac{2.d.t+d}{2.d.t-d}=\frac{d\left( 2t+1 \right)}{d\left( 2t-1 \right)}=\frac{2t+1}{2t-1}$
$\Rightarrow \frac{2a+b}{2a-b}=\frac{2c+d}{2c-d}$
c) $\frac{{{a}^{2}}+{{c}^{2}}}{{{b}^{2}}+{{d}^{2}}}=\frac{{{\left( b.t \right)}^{2}}+{{\left( d.t \right)}^{2}}}{{{b}^{2}}+{{d}^{2}}}=\frac{{{t}^{2}}\left( {{b}^{2}}+{{d}^{2}} \right)}{{{b}^{2}}+{{d}^{2}}}={{t}^{2}}$
$\frac{{{a}^{2}}-{{c}^{2}}}{{{b}^{2}}-{{d}^{2}}}=\frac{{{\left( b.t \right)}^{2}}-{{\left( d.t \right)}^{2}}}{{{b}^{2}}-{{d}^{2}}}=\frac{{{t}^{2}}\left( {{b}^{2}}-{{d}^{2}} \right)}{{{b}^{2}}-{{d}^{2}}}={{t}^{2}}$
$\Rightarrow \frac{{{a}^{2}}+{{c}^{2}}}{{{b}^{2}}+{{d}^{2}}}=\frac{{{a}^{2}}-{{c}^{2}}}{{{b}^{2}}-{{d}^{2}}}$
d) $\frac{a}{b}=\frac{c}{d}\Rightarrow \frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\Rightarrow {{\left( \frac{a-b}{c-d} \right)}^{2}}=\frac{a}{c}.\frac{b}{d}=\frac{ab}{cd}$