Let $[AD]$ and $[CE]$ be internal angle bisectors of $\triangle ABC$ such that $D$ is on $[BC]$ and $E$ is on $[AB]$. Let $K$ and $M$ be the feet of perpendiculars from $B$ to the lines $AD$ and $CE$, respectively. If $|BK|=|BM|$, show that $\triangle ABC$ is isosceles.