Question

6. AM is a median of a triangle BC . Show that
AB + BC + CA > 2 AM​

Answer 1

Step-by-step explanation:

AM is a median. So, BM = CM

CONSTRUCTION: Extend AM to D, such that AM= MD

=> ABDC is a parallelogram ( as diagonals are bisecting each other)

Since, AB + BM > AM……………..(1) ( The sum of 2 sides of a triangle > third side)

& BD + BM > MD ………….(2) ( the same reason)

Now, by adding (1) & (2)

We get, AB + BD + 2 BM > AM + MD ………(3)

But BD = AC ( opposite sides of parallelogram)

& 2BM = BC

& AM = MD

SO, Eq (3) becomes

AB + AC + BC = 2AM

[ proved]

HOPE IT HELPED YOU!!

Answer 2

since we know that the sum of two sides of a triangle is always greater than the third side,

In △ABM
AB+BM>AM ——-(i)


In △AMC
AC+MC>AM ——-(ii)


on adding equations (i) and (ii),
we have,

(AB+BM)+(AC+MC)>AM+AM
⇒AB+(BM+MC)+AC>2AM
⇒AB+BC+AC>2AB

Hence, proved.