How to prove that the median and angle bisector are not the same for non-isoceles triangles.

Question

I can prove that if the median and angle bisector are the same, then the triangle must be isosceles.

How can I prove that the median and angle bisector are not the same for non-isoceles triangles?

Assume triangle ABC with median AD and angle bisector AE.

Thanks in advance.

Answer 1

Using known expressions for median $m_a$ and corresponding bisector $\beta_a$

\begin{align} m_a^2 &= \tfrac12 b^2+\tfrac12 c^2-\tfrac14 a^2 ,\\ \beta_a^2 &= bc \left(1-\frac{a^2}{(b+c)^2}\right) , \end{align}

to get them of the same length, we must have \begin{align} \tfrac12 b^2+\tfrac12 c^2-\tfrac14 a^2 - bc \left(1-\frac{a^2}{(b+c)^2}\right) &=0 ,\\ \frac{(b-c)^2(2(b+c)^2-a^2)}{4(b+c)^2}&=0 , \end{align}

which is true only when either $a=\sqrt2(b+c)$, which never holds for triangles, or $b=c$.

Answer 2

The median $AI$ issued from vertex $A$ divides the opposite side $BC$ in two equal parts: $$\dfrac{IB}{IC}=1$$

The angle bisector $AJ$ of angle $\hat{A}$ divides the opposite side $BC$ into two segments whose lengths are in the same ratio as the adjacent sidelengths: $$\dfrac{JB}{JC}=\dfrac{AB}{AC}$$

(Angle bissector theorem)

The conclusion is easy from there.