show that $LM \perp LN$ and $LM = LN$ Consider $\Delta ABC$. L is a midpoint of side $BC$ Squares AFGB and ACDE are drawn exterior to $\Delta ABC$
Let M be a midpoint of diagonal of $AFGB$ and N be a midpoint of diagonal of $ACDE$ respectively. 
I need to show that $LM \perp LN$ and $LM = LN$  
My Attempt : As nothing about triangle ABC is mentioned I am confused. I made a construction (Unable to upload here) 
Is it that Apollonius' theorem which will help ? 
I am stuck 

You can convert this to an equivalent problem by backtracking: Given a square $MLNP$ and an arbitrary point $A$, show that $A$ rotated clockwise by 90 degrees around $M$ (denoted $B$) and $A$ rotated counterclockwise by 90 degrees around $N$ (denoted $C$) form a line segment such that $L$ is the midpoint of $BC$. I think it's not hard from there, by a few possible methods.
Reworded: Complete $MN$ to a square $ML'NP$. Clearly $A$ rotated around $N$ by 90 degrees becomes $C$ and the same rotated around $M$ by 90 degrees becomes $B$. The triangle $NPA$ is equivalent to $NL'C$ and the triangle $MPA$ is equivalent to $ML'B$ and $\vec{L'B} = -\vec{L'C}$, so $L'$ is the midpoint of $BC$ and thus $L' = L$. But for $L'$ the claim to be proved follows trivially.
Picture (hand-drawn in a short work break and shot on a cellphone, anyone feel free to replace by something better looking):

