如图,四边形$ABCD$与四边形$EFGH$位似,其位似中心为点$O$,且$frac{OE}{EA}=frac{4}{3}$,则$frac{FG}{BC}=$
$because$四边形$ABCD$与四边形$EFGH$位似,$thereforetriangleOEFsimtriangleOAB,triangleOFGbacksimtriangleOBC$,$thereforefrac{OF}{OB}=frac{OE}{OA}=frac{4}{7},quadthereforefrac{FG}{BC}=frac{OF}{OB}=frac{4}{7}$.