Prove that the ratio of the perimeters of two similar triangles is the same as the ratio of their corresponding sides.

Let two similar triangles be △ABC and △PQR.

We know that when triangles are similar ratio of corresponding sides are equal.

$\therefore \dfrac{AB}{PQ} = \dfrac{BC}{QR} = \dfrac{AC}{PR}.$

By property of ratio i.e.,

if $\dfrac{a}{b} = \dfrac{b}{c} = \dfrac{d}{e},$ then each ratio = $\dfrac{\text{sum of antecedents}}{\text{sum of consequents}}$.

So,

$\dfrac{AB}{PQ} = \dfrac{BC}{QR} = \dfrac{AC}{PR} = \dfrac{AB + BC + AC}{PQ + QR + PR}$

Since, AB + BC + AC = Perimeter of △ABC and PQ + QR + PR = Perimeter of △PQR. So,

$\dfrac{AB}{PQ} = \dfrac{BC}{QR} = \dfrac{AC}{PR} = \dfrac{\text{Perimeter of △ABC}}{\text{Perimeter of △PQR}}.$

Hence, proved.