Question

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

Answer 1

Let the two similar triagles be ABC and PQR and angles A = P, B = Q and C = R.


In similar triangles the sides opposite to equal angles are proportional.

Hence, $\frac{BC}{QR}=\frac{AC}{PR}=\frac{AB}{PQ}=k$, (constant of proportionality)
From the property of proportion, if $\frac{a}{b}=\frac{c}{d}=\frac{e}{f}$, then $\frac{(a+b+c)}{(p+q+r)}=\frac{a}{b}=\frac{c}{d}=\frac{e}{f}=k$
Hence $\frac{(AB+BC+CA)}{PQ+QR+RP}=\frac{perimeteroftriangleABC}{perimeteroftrianglePQR}=k$
Proved