the length of a perpendicular inside of a triangle Let $ABC$ a triangle with its area $S$ and $AB=c$, $BC=a$, $AC=b$.
The median from $A$ intersects the bisecting of the angle $ABC$ in $X$.
If $x=\ $the length of the perpendicular from $X$ on $BC$ than $x$ is:
You are almost there: if $H$ is the perpendicular projection from $A$ on $BC$ and $Q$ is the perpendicular projection from $X$ on $BC$ ($XQ=x$), then:
$$
AH/XQ=AM/XM=1+AX/XM
\quad\text{and}\quad
AH=2S/a.
$$
