Chapter 6.4, Problem 21WE

To determine

To Prove: The sum of the lengths of the medians is greater than half the perimeter of the triangle.

Explanation of Solution

Given information:

  $\begin{array}{l}AB+BC>AC\hfill \\ AB+AC>BC\hfill \\ AC+BC>AB\hfill \end{array}$

The sum of the lengths of any two sides of a triangle is greater than the length of the third side.

  $\begin{array}{l}AB+BC>AC\hfill \\ AB+AC>BC\hfill \\ AC+BC>AB\hfill \end{array}$

  

So the figure can be drawn as shown below:

  

Apply above theorem,

In ΔACP,

  $AP+CP>AC\mathrm{........}\left(1\right)$

Similarly, in ΔABQ,

  $AQ+QB>AB\mathrm{.......}\left(2\right)$

And, in ΔBRC,

  $BR+RC>BC\mathrm{.........}\left(3\right)$

Perimeter of a triangle is $AB+BC+CA$

Sum of the lengths of the medians is $AQ+CP+BR$

Add equations (1), (2) and (3) to get

  $\begin{array}{l}AP+CP+AQ+QB+BR+RC>\left(AC+AB+BC\right)\hfill \\ AQ+CP+BR>\left(AC+AB+BC\right)–AP–QB–RC\hfill \\ AQ+CP+BR>\left(\left(AC–RC\right)+\left(AB–AP\right)+\left(BC–QB\right)\right)\hfill \end{array}$

As the value of $AR=RC=\frac{AC}{2}$ Similarly,

  $\begin{array}{l}AP=PB\hfill \\ =\frac{AB}{2}\hfill \end{array}$

And

  $\begin{array}{l}BQ=QC\hfill \\ =\frac{BC}{2}\hfill \end{array}$

Therefore,

  $\begin{array}{l}AQ+CP+BR>\left(\frac{AC}{2}+\frac{AB}{2}+\frac{BC}{2}\right)\hfill \\ AQ+CP+BR>\left(\frac{AC+AB+BC}{2}\right)\hfill \end{array}$

Therefore, sum of the lengths of the medians is greater than half the perimeter of the triangle.

Hence, $AQ+CP+BR>\left(\frac{AC+AB+BC}{2}\right)$