in a triangle abc am is a median of abc and hence bm =mc. Proove that ab+bc+ca>2am
In any triangle, the sum of any two sides > the third side
so in triangle ABM:
AB + BM > AM
in triangle ACM
AC + MC > AM
adding:
AB + AC + BM+MC > 2AM
AB + AC + BC >2AM
btw, this is true not just for the median AM
In fact, the same steps can be used to prove it true if M is any point on BC.
To prove that `ab + bc + ca > 2am` using the given information that `am` is the median of triangle ABC, we can start by using the triangle inequality theorem.
The triangle inequality theorem states that for any triangle, the sum of the lengths of any two sides of the triangle must be greater than the length of the remaining side.
In triangle ABC, since `am` is a median, it divides side BC into two equal parts, making `bm` and `mc` equal in length. Therefore, we have `bm = mc`.
Now, let's divide the inequality `ab + bc + ca > 2am` into two separate inequalities:
1. `ab + am > bm`
2. `ca + am > mc`
Substituting `bm = mc`, we get:
1. `ab + am > mc`
2. `ca + am > mc`
Now, let's prove each inequality separately:
1. To prove `ab + am > mc`, we can use the triangle inequality theorem on triangle AMC:
- `ac + am > mc`
- By substitution, we have `ca + am > mc`
2. To prove `ca + am > mc`, we can use the triangle inequality theorem on triangle AMB:
- `ab + am > bm`
- By substitution, we have `ab + am > mc`
Since both `ab + am > mc` and `ca + am > mc` are true, we can combine them to get:
`ab + am + ca + am > mc + mc`
`ab + ca + 2am > 2mc`
As we know that `bm = mc`, we can substitute `bm` with `mc`:
`ab + ca + 2am > 2bm`
Finally, since `ab + bc + ca` is always greater than `ab + ca`, we can conclude:
`ab + bc + ca > ab + ca + 2am > 2bm`
Therefore, we have proved that `ab + bc + ca > 2am` using the fact that `am` is the median of triangle ABC.