Question

In \[\triangle ABC\] and \[\triangle DEF\], \[\angle A = \angle D\], \[\angle B = \angle E\] and \[\angle C = \angle F.\] Also, \[AL\] and \[DM\] are medians. Prove that \[\frac{BC}{EF} = \frac{AL}{DM}.\]

Answer 1

In \[\triangle ABC\] and \[\triangle DEF\],

⇒ \[\angle A = \angle D\] (Given)

⇒ \[\angle B = \angle E\] (Given)

⇒ \[\angle C = \angle F\] (Given)

∴ \[\triangle ABC \sim \triangle DEF\] (By AAA postulate)

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

∴ \[BL = \frac{1}{2}\, BC\] and \[EM = \frac{1}{2}\, EF\]

We know that,

Corresponding sides of similar triangles are proportional.

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

⇒ \[\frac{AB}{DE} = \frac{\frac{1}{2}\, BC}{\frac{1}{2}\, EF}\]

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

In \[\triangle ABL\] and \[\triangle DEM\],

⇒ \[\angle B = \angle E\] (Given)

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

∴ \[\triangle ABL \sim \triangle DEM\] (By SAS postulates)

As, corresponding sides of similar triangles are proportional.

∴ \[\frac{AB}{DE} = \frac{AL}{DM} \qquad \text{.........(3)}\]

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

⇒ \[\frac{BC}{EF} = \frac{AL}{DM}\]

Hence, proved that \[\frac{BC}{EF} = \frac{AL}{DM}\]