$证明：∵△ABC\sim △A'B'C'$
$∴∠B=∠B'，\frac {AB}{A'B'}=\frac {BC}{B'C'}=k$
$∵AD与A'D'分别是△ABC和△A'B'C'中边BC、B'C'上的中线$
$∴BC= 2BD，B'C'=2B'D'$
$∵\frac {BC}{B'C'}=k$
$∴\frac {2BD}{2B'D'}=k$
$∴\frac {BD}{B'D'}=k$
$∴\frac {AB}{A'B'}=\frac {BD}{B'D'}=k$
$∴△ABD∽△A'B'D'$
$∴\frac {AB}{A'B'}=\frac {AD}{A'D'}=k$
