Question

In triangleABC, D is any point on BC.
Prove that : AB + BC + AC > 2 AD.​

Answer 1

Step-by-step explanation:

Clearly AB+BD>AD -1

and, AC+CD>AD-2

adding equation 1and2

AB + BD +AC+ CD>AD+AD or 2AD

Answer 2

$\huge\mathcal{Answer:}$

In a triangle sum of two sides is always greater than the third side.

In ABD

$ab + bd > ad - - (1)$

InADC

$dc + ca > ad - - (2)$

on adding (1) and (2)

$ab + bd + dc + ca > 2ad$

on putting BD+Dc=BC

$ab + bc +ca > 2ad$

$\huge\mathcal{Hence proved♡}$