Question

(2) AM is a median of a triangle - ABC. Prove that
AB + BC + CA > 2 AM​

Answer 1

$\huge\bold\green{To\:proof:}$

AB + BC + CA > 2AM

$\huge\bold\green{proof:}$

LET M BE THE POINT ON BC WHERE DM INTERSECTS BC

Consider TRIANGLE AMB

AB + BM > AM (Sum of opposite side of a triangle is greater than third side) ----(i)

Consider TRIANGLE AMC

AC + CM > AM (Sum of opposite side of a triangle is greater than third side) ----(ii)

From i and ii

AB + BM + AC + CM > AM + AM

AB + BC + AC > 2AM

HENCE PROVED