Question:

Proof (NCERT method)

1. Given two triangles \(\triangle ABC\) and \(\triangle DEF\) such that
$$\frac{AB}{DE}=\frac{BC}{EF}=\frac{CA}{FD}=k\;(k>0).$$
2. Construct a triangle \(\triangle A'B'C'\) on the same plane as \(\triangle ABC\) by enlarging \(\triangle ABC\) by the factor \(k\). Thus
\[A'B'=k\cdot AB,\; B'C'=k\cdot BC,\; C'A'=k\cdot CA.\]
By the given proportion, these lengths are exactly the lengths of \(DE, EF, FD\):
\[A'B'=DE,\; B'C'=EF,\; C'A'=FD.\]
3. Apply SSS congruence: Since \(\triangle A'B'C'\) and \(\triangle DEF\) have all three corresponding sides equal, they are congruent (SSS). Hence their corresponding angles are equal:
\[\angle A'B'C' = \angle D E F,\; \angle B'C'A' = \angle E F D,\; \angle C'A'B' = \angle F D E.\]
4. Relate back to the original triangles: Because \(\triangle A'B'C'\) is just a scaled‑up version of \(\triangle ABC\), the angles of \(\triangle A'B'C'\) are the same as those of \(\triangle ABC\). Therefore
\[\angle ABC = \angle DEF,\; \angle BCA = \angle EFD,\; \angle CAB = \angle FDE.\]
5. Conclusion: All three corresponding angles are equal, so the two triangles are similar. Hence, if the three sides of one triangle are in the same ratio as the three sides of another triangle, the triangles are similar – the SSS similarity criterion.

Answer: The statement is true; the triangles are similar by the SSS similarity criterion.