Question

AM is a median of ABC. Provec that AB + BC+ CA > 2 AM​

Answer 1

Step-by-step explanation:

As we know that the sum of lengths of any two sides should be greater than the length of third side.

therefore,

In ∆ ABM

AB + BM > AM .......... ( i )

In ∆ AMC

AC + MC > AM ............. ( ii )

Adding equ. ( i ) & ( ii ) we have,

( AB + BM ) + ( AC + MC ) > AM + AM

= AB + ( BM + MC ) + AC > 2 AM

= AB + BC + AC > 2 AM

HENCE PROVED