Prove that the ratio of the perimeters of two similar triangles is the same as the

Let the two triangles be ABC and PQR.
We have:

$\triangle A B C \sim \triangle P Q R$

Here,
BC = a, AC = b and AB = c
PQ = r, PR = q and QR = p

We have to prove:

$\frac{a}{p}=\frac{b}{q}=\frac{c}{r}=\frac{a+b+c}{p+q+r}$

$\triangle A B C \sim \triangle P Q R$; therefore, their corresponding sides will be proportional.

$\Rightarrow \frac{a}{p}=\frac{b}{q}=\frac{c}{r}=k$ (say) $\quad \ldots$ (i)

$\Rightarrow a=k p, b=k q$ and $c=k r$

$\therefore \frac{\text { Perimeter of } \triangle A B C}{\text { Perimeter of } \triangle P Q R}=\frac{a+b+c}{p+q+r}=\frac{k p+k q+k r}{p+q+r}=k \quad \ldots$ (ii)

From (i) and (ii), we get:

$\frac{a}{p}=\frac{b}{q}=\frac{c}{r}=\frac{a+b+c}{p+q+r}=\frac{\text { Perimeter of } \triangle A B C}{\text { Perimeter of } \triangle P Q R}$

This completes the proof.