Đặt $\dfrac ab =\dfrac cd = k$
$\Rightarrow \begin{cases}a = kb\\c = kd\end{cases}$
Ta được:
$+)\quad (a+2c)(b+d)$
$= (kb + 2kd)(b+d)$
$= k(b+2d)(b+d)\quad (1)$
$+)\quad (a+c)(b+2d)$
$= (kb + kd)(b+2d)$
$= k(b+d)(b+2d)\quad (2)$
Từ $(1)(2)\Rightarrow (a+2c)(b+d)= (a+c)(b+2d)$