Question

In the given figure, \(\triangle \mathrm{ABC}\) and \(\triangle \mathrm{DEF}\) are similar. BM and EN are their medians. Prove that:

1. \[\triangle AMB \sim \triangle DNE\]
2. \[\triangle CMB \sim \triangle FNE\]
3. \[\frac{BM}{EN} = \frac{AC}{DF}\]

Answer 1

(i) Given,

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

∴ \[AM = \frac{1}{2}\, AC\] and \[DN = \frac{1}{2}\, DF\]

Given,

\[\triangle ABC \sim \triangle DEF\]

∴ \[\angle A = \angle D\] (Corresponding angles of similar triangles are equal)

We know that,

Corresponding sides of similar triangles are proportional.

∴ \[\frac{AB}{DE} = \frac{AC}{DF} \qquad \text{....(1)}\]

⇒ \[\frac{AB}{DE} = \frac{\frac{1}{2}\, AC}{\frac{1}{2}\, DF}\]

⇒ \[\frac{AB}{DE} = \frac{AM}{DN} \qquad \text{....(2)}\]

In \[\triangle AMB\] and \[\triangle DNE\],

⇒ \[\angle A = \angle D\] (Proved above)

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

∴ \[\triangle AMB \sim \triangle DNE\] (By SAS postulate)

(ii) Given,

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

∴ \[MC = \frac{1}{2}\, AC\] and \[NF = \frac{1}{2}\, DF\]

Given,

\[\triangle ABC \sim \triangle DEF\]

⇒ \[\angle C = \angle F\] (Corresponding angles of similar triangles are equal)

Since, corresponding sides of similar triangles are proportional.

∴ \[\frac{BC}{EF} = \frac{AC}{DF} \qquad \text{....(3)}\]

⇒ \[\frac{BC}{EF} = \frac{\frac{1}{2}\, AC}{\frac{1}{2}\, DF}\]

⇒ \[\frac{BC}{EF} = \frac{MC}{NF} \qquad \text{....(4)}\]

In \[\triangle CMB\] and \[\triangle FNE\],

⇒ \[\angle C = \angle F\] (Proved above)

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

∴ \[\triangle CMB \sim \triangle FNE\] (By SAS postulate)

(iii) Given,

\[\triangle ABC \sim \triangle DEF\]

We know that,

Corresponding sides of similar triangles are proportional.

∴ \[\frac{AB}{DE} = \frac{AC}{DF} \qquad \text{.......(5)}\]

\[\triangle AMB \sim \triangle DNE\]

We know that,

Corresponding sides of similar triangles are proportional.

∴ \[\frac{BM}{EN} = \frac{AB}{DE} \qquad \text{......(6)}\]

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

⇒ \[\frac{BM}{EN} = \frac{AC}{DF}\]

Hence, proved that \[\frac{BM}{EN} = \frac{AC}{DF}.\]