Relationship between the altitude of an isosceles triangle and segments drawn to the lateral side from a point on the base. Question :In an isosceles triangle, the sum of the distances from each point of the base to the lateral sides is constant.
I've tried a couple of things, but it seems like this statement is not true. Can someone provide a HINT towards a solution of the problem, if any? I don't want an answer I just want a minor hint that will guide me.

Hint: Suppose you have an isosceles triangle $\Delta ABC$ with base $\overline{BC}$. Then you can draw a second isosceles triangle $\Delta A'BC$ by reflecting $A'$ across $\overline{BC}$, so that $ABA'C$ is a parallelogram. Now draw a line which intersects $\overline{BC}$ and is perpendicular to $AB$. This line is divided into two segments by $\overline{AB},\overline{BC},$ and $\overline{A'B}$. Can you see a relationship between these segments and your question?
