Let $G$ be the intersecion of the perpendicular from $B$ to $AC$ with $AC$. In my proof I asume $G$ lies between $F$ and $L$. Since $AB^2+BC^2 = AL^2+LC^2$ we have $BG^2+AG^2+BG^2+GC^2 = AL^2+LC^2$. Therefore: $2BG^2 = (AL-AG)(AL+AG)+(LC-GC)(LC+GC)$ $2BG^2= GL (AL+AG) - GL(LC+GC)$ $2BG^2 = GL(AL+AG-LC-GC) $ $=> 2BG^2 = GL(AF+FG+GL+AF+FG-LC - GL-LC) = 2GL \cdot FG$. Hence $BG^2= GL \cdot FG$ which imples that $\angle FBL = 90$ (because of the altitude in the hypotenuse formulas)

Using the scalar product, we get : \begin{align*}\overrightarrow{BF}\cdot\overrightarrow{BL}&=(\overrightarrow{BA}+\overrightarrow{AF})\cdot(\overrightarrow{BC}+\overrightarrow{CL})\\&=(\overrightarrow{BA}+\overrightarrow{LC})\cdot(\overrightarrow{BC}-\overrightarrow{LC})\\&=\overrightarrow{BA}\cdot\overrightarrow{BC}+\overrightarrow{LC}\cdot(\overrightarrow{BC}-\overrightarrow{BA})-LC^2\\&=\frac{AB^2+BC^2-AC^2}{2}+\overrightarrow{LC}\cdot\overrightarrow{AC}-LC^2\\&=\frac{AB^2+BC^2-(AL+LC)^2+2LC(AC-LC)}{2}\\&=\frac{(AL^2+LC^2)-(AL^2+LC^2 + 2LC \cdot AL)+2LC\cdotAL}{2}\\&=0\end{align*} Q.E.D.

Say $A=(a,0)$, $F=(1,0)$ and $B(x,y)$. Then $C= (-a,0)$ and $L(-1,0)$. So we have: $$ (x+a)^2+y^2+(x-a)^2+y^2 = (a-1)^2+(a+1)^2$$ from where we get $x^2+y^2=1$. So $B$ is on circle with diameter $LY$ and thus $\angle LBF = 90^{\circ}$.

An easy geometry....... Follows