Prove that the perimeter of a triangle is greater than the sum of its altitudes.

To do:

We have to prove that the perimeter of a triangle is greater than the sum of its altitudes.

Solution:

Let in $\triangle ABC$,

$AD, BE$ and $CF$ are altitudes.

We know that, 

Side opposite to the greatest angle is the largest.

In $\triangle ABD$,

$\angle D = 90^o$

This implies,

$\angle D > \angle B$

Therefore,

$AB >AD$.....…(i)

Similarly,

$BC > BE$.........(ii)

$CA > CF$.........(iii)

Adding (i), (ii) and (iii), we get,

$AB + BC + CA > AD + BE + CF$

Therefore, the perimeter of a triangle is greater than the sum of its altitudes.

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Updated on: 10-Oct-2022

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