Question

Prove that the sum of three medians of a triangle is less than the perimeter of the triangle.

Answer 1

Given:

• Triangle ABC with medians AD, BE, CF (D, E, F are midpoints of BC, CA, AB respectively).

To Prove:

• AD + BE + CF < AB + BC + CA

Proof [Step-wise]:

1. Show `AD < (AB + AC)/2`.

Extend AD to a point E on the line AD beyond D so that DE = AD.

Then D is the midpoint of AA’, where A’ is called E in the construction is the point on the extension and AA’ = 2·AD.

By construction and congruence (see Example 8), one obtains AB = EC. 

So, EC = AB. 

In triangle AEC, by the triangle inequality, EC + AC > AE.

Substituting EC = AB and AE = 2·AD gives AB + AC > 2·AD.

Hence, `AD < (AB + AC)/2`.

2. By identical reasoning cyclic permutation of vertices, we get

`BE < (BC + BA)/2`

`CF < (CA + CB)/2`

3. Add the three inequalities:

`AD + BE + CF < (AB + AC)/2 + (BC + BA)/2 + (CA + CB)/2`

= `(2AB + 2BC + 2CA)/2`

= AB + BC + CA

Therefore, AD + BE + CF < AB + BC + CA.

The sum of the three medians of triangle ABC is less than the perimeter of the triangle.