Santa draws a Christmas tree in the following way. He first draws an acute-angled triangle $\vartriangle ABC$. He then lets $M$ be the mid-point of $BC$, $A'$ be the point of reflection of $A$ over $BC$, $D$ be the point of intersection of line segment $AM$ and the circumcircle of $\vartriangle A'BC$, $E$ and $F$ be points on $AB$ and $AC$ respectively such that $D$, $E$, $F$ are collinear. Prove that $\frac{AE}{ED \cdot DB} = \frac{AF}{FD\cdot DC}$ . Note: Line segments $AE$, $ED$, $DB$, $BC$, $CD$, $DF$, $FA$ form the shape of a Christmas tree!