Question 13 Marks
$AM$ is a median of a triangle $ABC.$

Is $AB + BC + CA > 2 AM?($Consider the sides of triangles $\triangle ABM$ and $\triangle AMC.)$

Is $AB + BC + CA > 2 AM?($Consider the sides of triangles $\triangle ABM$ and $\triangle AMC.)$
Answer
View full question & answer→In $\triangle ABM$
$AB + BM > AM....[$Sum of the lengths of any two sides of a triangle is greater than the length of the third side$].....(1)$
In $\triangle ACM$
$CA + CM > AM....[$Sum of the lenghts of any two sides of a triangle is greater than the length of the third side$].....(2)$
Sum $(1)$ and $(2)$
$(AB + BM) + (CA + CM) > AM + AM$
$ \therefore AB + (BM + CM) + CA > 2AM$
$\therefore AB + BC + CA > 2AM$
$AB + BM > AM....[$Sum of the lengths of any two sides of a triangle is greater than the length of the third side$].....(1)$
In $\triangle ACM$
$CA + CM > AM....[$Sum of the lenghts of any two sides of a triangle is greater than the length of the third side$].....(2)$
Sum $(1)$ and $(2)$
$(AB + BM) + (CA + CM) > AM + AM$
$ \therefore AB + (BM + CM) + CA > 2AM$
$\therefore AB + BC + CA > 2AM$















