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