1
$\begingroup$

Prove that the median of the largest segment is the smallest.

In the book it is written that it is easy provable using triangle inequality But where should I perform the triangle inequality?Any hints?

  • 1
    You should say: "Prove that, in a triangle, the median associated with the largest side has the smallest length" – 2017-01-26
  • 0
    @JeanMarie Then how to prove? – 2017-01-26
  • 0
    @ChristianBlatter As JeanMarie said. – 2017-01-26

2 Answers 2

2

Let the centroid of the triangle be at the origin, and let ${\bf a}$, ${\bf b}$, ${\bf c}$ be the position vectors of its vertices $A$, $B$, $C$. Then $${\bf a}+{\bf b}+{\bf c}={\bf 0}\ ;\tag{1}$$ furthermore the lengths of the medians are ${3\over2}|{\bf a}|$, ${3\over2}|{\bf b}|$, ${3\over2}|{\bf c}|$.

Assume that $|AB|\geq|AC|$, which is the same as $|{\bf b}-{\bf a}|\geq|{\bf c}-{\bf a}|$. Eliminating ${\bf a}$ here using $(1)$ gives $|2{\bf b}+{\bf c}|\geq|{\bf b}+2{\bf c}|$, and by squaring we obtain $$4|{\bf b}|^2+4{\bf b}\cdot{\bf c}+|{\bf c}|^2\geq |{\bf b}|^2+4{\bf b}\cdot{\bf c}+4|{\bf c}|^2\ .$$ This implies $|{\bf b}|\geq|{\bf c}|$, as claimed.

2

I assume that $a \lt b \lt c$.

enter image description here

Use cosine law twice to setup equations for

(1) … $(M_C)^2$ in terms of b, c, cos A.

(2) …$(M_B)^2$ in terms of b, c, cos A.

In the process of doing (1) – (2), the terms involving cosine will be cancelled out. From the assumed fact that $c \gt b$, we can arrive at the conclusion of $M_B \gt M_C$.

Note:- Like to see a solution using triangle inequality.