$\frac{a+b}{c+b}=\frac{c+d}{a+d}\Rightarrow a=c$ or $a+b+c+d=0$ Please find my mistake or give hint:
If $\frac{c}{a}\geq 1$, then by mediant inequality:
$1\leq\frac{a+b}{c+b}$ and $\frac{c+d}{a+d}\leq 1\Rightarrow c=a$
If $\frac{a}{c}\geq 1$, then:
$\frac{a+b}{c+b}\leq 1$ and $1\leq \frac{c+d}{a+d}\Rightarrow c=a$
thnks

Note that
$$
\frac{a+b}{c+b}  = \frac{c+d}{a+d} \implies
(a+d)(a+b) = (c+d)(c+b)\implies\\
a^2 + ab + ad + db = c^2 + cb + cd + db \implies\\
a^2 - c^2 +ab - cb + ad - cd = 0 \implies\\
(a-c)(a+c) + (a-c)b + (a-c)d = 0 \implies\\
(a-c)(a+b+c+d) = 0
$$
