Three equal cevians In a scalene triangle,does there exist three cevians which are equal in length,where length is measured between the corresponding vertex and the intersection point of the cevian with the corresponding side? This is just a question I had in my mind. 
 A: Here is a bit of a trivial answer. Let the sides of the triangle be $a\le b\le c$, and choose cevians as the side $AC$ from $A$, the side $CA$ from $C$ (both of length $b$). Then the fact that we chose the side $b$ as the middle side means that there is a point $D$ on $AC$ so that the cevian $BD$ has length $b$.
A: Let $a$, $b$ and $c$ be the sides of the triangle, $a<b<c$. Let $h_a>h_b>h_c$ be the heights.
The length of a cevian from the vertex $X$ is between two sides $y$ and $z$ if the height $h_x$ is outside the triangle, or between the height and the longest side if the height is inside. 
Suppose for now that the three angles are acute and the three heights are inside. Then, there will be three equal cevians if and only if
$$I_a\cap I_b\cap I_c=[h_a,c)\cap[h_b,c)\cap[h_c,b)$$
is empty.
Since $h_a>h_b$, $I_a\subset I_b$, thus, the intersection is $I_a\cap I_c$, which is empty if and only if
$$b\leq h_a$$
but $h_a=b\sin \hat C$, and since $\hat C$ is acute, this is impossible.
If the triangle is right angled or has an obtuse angle,
$$I_a\cap I_b\cap I_c=(b,c)\cap (a,c)\cap [h_a,b)=\emptyset$$
So the answer is "yes" if and only if the three angles are acute.
Remark: if you consider that a side is a cevian, then right angled triangles also has three equal cevians of length $b$ (actually two of them are the longest leg).
