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AD is the altitude of the triangle corresponding to side BC. Prove that $ A B+A C+B C>2 A D $.
Given: ABC is a triangle.
AD is the altitude corresponding to BC.
To do:
We have to prove $AB+AC+BC>2AD$.
Solution:
In triangle ABD,
$AB+BD>AD$ (Sum of any two sides of a triangle is greater than the third side)...(i)
In triangle ACD,
$DC+AC>AD$.........(ii)
Adding equations (i) and (ii), we get,
$AB+(BD+DC)+AC>AD+AD$
$AB+AC+BC>2AD$
Hence proved.
Updated on: 10-Oct-2022
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