Đặt $\dfrac ab =\dfrac cd = k$
$\to \begin{cases}a = kb\\c = kd\end{cases}$
Ta có:
$+)\quad \dfrac{a^2 + c^2}{b^2 + d^2}$
$=\dfrac{k^2b^2 + k^2d^2}{b^2 + d^2}$
$= \dfrac{k^2(b^2 + d^2)}{b^2 + d^2}$
$= k^2$
$+)\quad \dfrac{(a + c)^2}{(b+d)^2}$
$=\dfrac{(kb + kd)^2}{(b+d)^2}$
$= \dfrac{k^2(b + d)^2}{(b + d)^2}$
$= k^2$
Do đó:
$\dfrac{a^2 + c^2}{b^2 + d^2} =\dfrac{(a + c)^2}{(b+d)^2}$