In triangle ABC and DEF, ∠A = ∠D, ∠B = ∠E and ∠C = ∠F. Also, AL and DM are medians. Prove that $\dfrac{\text{BC}}{\text{EF}} = \dfrac{\text{AL}}{\text{DM}}$.

In Δ ABC and Δ DEF,

⇒ ∠A = ∠D (Given)

⇒ ∠B = ∠E (Given)

⇒ ∠C = ∠F (Given)

∴ Δ ABC ∼ Δ DEF (By AAA postulate)

Since, AL and DM are medians of triangles ABC and DEF respectively.

∴ BL = $\dfrac{1}{2}BC$ and EM = $\dfrac{1}{2}EF$

We know that,

Corresponding sides of similar triangles are proportional.

$\therefore \dfrac{AB}{DE} = \dfrac{BC}{EF}………………..(1)\\[1em] \Rightarrow \dfrac{AB}{DE} = \dfrac{\dfrac{1}{2}BC}{\dfrac{1}{2}EF}\\[1em] \Rightarrow \dfrac{AB}{DE} = \dfrac{BL}{EM} …………………….(2)$

In Δ ABL and Δ DEM,

⇒ ∠B = ∠E (Given)

⇒ $\dfrac{AB}{DE} = \dfrac{BL}{EM}$ [From equation (2)]

∴ Δ ABL ∼ Δ DEM (By SAS postulates)

As, corresponding sides of similar triangles are proportional.

$\therefore \dfrac{AB}{DE} = \dfrac{AL}{DM}$ ………(3)

From equation (1) and (3), we get :

$\Rightarrow \dfrac{\text{BC}}{\text{EF}} = \dfrac{\text{AL}}{\text{DM}}$

Hence, proved that $\dfrac{\text{BC}}{\text{EF}} = \dfrac{\text{AL}}{\text{DM}}$