Áp dụng tính chất tỉ lệ thức
$→(a+3c)(b+d)=(b+3d)(a+c)$
$[a(b+d)]+[3c(b+d)]=[b(a+c)]+[3d(a+c)]$
$(ab+ad)+(3cb+3cd)=(ab+bc)+(3ad+3cd)$
$ab+ad+3bc+3cd=ab+bc+3ad+3cd$
$ab+ad+3bc+3cd-ab-bc-3ad-3cd=0$
$(ab-ab)+(ad-3ad)+(3bc-bc)+(3cd-3cd)=0$
$-2ad+2bc=0$
$-2ad=-2bc$
$ad=bc$
$→\dfrac{a}{b}=\dfrac{c}{d}$
$→$ ĐPCM