In triangle ABC and DEF, ∠A = ∠D and $\dfrac{\text{AB}}{\text{AC}} = \dfrac{\text{DE}}{\text{DF}}$ then prove that Δ ABC ∼ Δ DEF.

Given, $\dfrac{\text{AB}}{\text{AC}} = \dfrac{\text{DE}}{\text{DF}}$

⇒ $\dfrac{\text{AB}}{\text{DE}} = \dfrac{\text{AC}}{\text{DF}}$

In Δ ABC and Δ DEF,

⇒ ∠A = ∠D (Given)

⇒ $\dfrac{\text{AB}}{\text{DE}} = \dfrac{\text{AC}}{\text{DF}}$

∴ Δ ABC ∼ Δ DEF (By SAS postulate)

Hence, proved that Δ ABC ∼ Δ DEF.