In the triangle $ABC$, $AB>AC$. Show that $BE>CF$ where $E$ and $F$ are the midpoints of $AC$ and $AB$

Question

In the triangle $ABC$, $AB>AC$ and $E,F$ are the midpoints of $AC$ and $AB$.
Show that $BE>CF$.

I going this problem in the excursion of mathematics book.

I try it very much but can't able to do it. Somebody help me .

Answer 1

By Stewart's theorem $$ 4m_b^2 = 2a^2+2c^2-b^2,\qquad 4m_c^2 = 2a^2+2b^2-c^2, $$ hence:

$$ 4(m_c^2 - m_b^2) = 3(b^2-c^2)$$

and $c>b$ is equivalent to $m_c < m_b$.