In the given figure, Δ ABC and Δ DEF are similar, BM and EN are their medians. If Δ ABC is similar to Δ DEF, prove that :

(i) Δ AMB ∼ Δ DNE

(ii) Δ CMB ∼ Δ FNE

(iii) $\dfrac{BM}{EN} = \dfrac{AC}{DF}$

(i) Given,

Since, BM and EN are medians of triangles ABC and DEF respectively.

∴ AM = $\dfrac{1}{2}AC$ and DN = $\dfrac{1}{2}DF$

Given,

Δ ABC ∼ Δ DEF

∴ ∠A = ∠D (Corresponding angles of similar triangles are equal)

We know that,

Corresponding sides of similar triangles are proportional.

$\therefore \dfrac{AB}{DE} = \dfrac{AC}{DF} ………………..(1)\\[1em] \Rightarrow \dfrac{AB}{DE} = \dfrac{\dfrac{1}{2}AC}{\dfrac{1}{2}DF}\\[1em] \Rightarrow \dfrac{AB}{DE} = \dfrac{AM}{DN} …………………….(2)$

In Δ AMB and Δ DNE,

⇒ ∠A = ∠D (Proved above)

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

∴ Δ AMB ∼ Δ DNE (By SAS postulate)

Hence, proved that Δ AMB ∼ Δ DNE.

(ii) Given,

Since, BM and EN are medians of triangles ABC and DEF respectively.

∴ MC = $\dfrac{1}{2}AC$ and NF = $\dfrac{1}{2}DF$

Given,

Δ ABC ∼ Δ DEF

⇒ ∠C = ∠F (Corresponding angles of similar triangles are equal)

Since, corresponding sides of similar triangles are proportional.

$\therefore \dfrac{BC}{EF} = \dfrac{AC}{DF} ………………..(3)\\[1em] \Rightarrow \dfrac{BC}{EF} = \dfrac{\dfrac{1}{2}AC}{\dfrac{1}{2}DF}\\[1em] \Rightarrow \dfrac{BC}{EF} = \dfrac{MC}{NF} …………………….(4)$

In Δ CMB and Δ FNE,

⇒ ∠C = ∠F (Proved above)

⇒ $\dfrac{BC}{EF} = \dfrac{MC}{NF}$ [From equation (4)]

∴ Δ CMB ∼ Δ FNE (By SAS postulate)

Hence, proved that Δ CMB ∼ Δ FNE.

(iii) Given,

Δ ABC ∼ Δ DEF

We know that,

Corresponding sides of similar triangles are proportional.

$\therefore \dfrac{AB}{DE} = \dfrac{AC}{DF}$ …….(5)

Δ AMB ∼ Δ DNE

We know that,

Corresponding sides of similar triangles are proportional.

$\therefore \dfrac{BM}{EN} = \dfrac{AB}{DE}$ ……(6)

From equation (5) and (6), we get :

$\Rightarrow \dfrac{BM}{EN} = \dfrac{AC}{DF}$

Hence, proved that $\dfrac{BM}{EN} = \dfrac{AC}{DF}$.