Question

Prove that ratio of their corresponding altitudes of two similar triangles is equal to ratio of their corresponding sides.

Answer 1

Let ∆ABC and ∆DEF are two similar triangles.

Given :- ∆ABC similar to triangle DEF, AL Perpendicular to BC and DM Perpendicular to EF

to prove:

Ar(∆ABC)/Ar(∆DEF) = AL²/DM^2

proof:

As we know that the ratio of the areas of two similar triangles is equal to the ratio of the squares of the corresponding sides.

Therefore,

Ar(∆ABC)/Ar(∆DEF) = AB²/DE² ........(1)

In ∆ALB and ∆DME , we have

Angle ALB = Angle DME = 90°

and, Angle B = Angle E { ∆ABC similar∆DEF)

Therefore,

∆ALB similar to ∆DME { By AA similarity}

=> AB/DE = AL/DM

=> AB²/DE² = AL²/DM² ........(2)

From 1 and 2 we get,

Ar(∆ABC)/Ar(∆DEF) = AL²/DM²..... PROVED.....