In the given figure,$OB$ is the perpendicular bisector of the line segment $DE$,$FA \perp OB$,and $FE$ intersects $OB$ at the point $C$. Prove that $\frac{1}{OA} + \frac{1}{OB} = \frac{2}{OC}$.

(N/A) In $\triangle AOF$ and $\triangle BOD$:
$\angle O = \angle O$ (Common angle) and $\angle A = \angle B = 90^{\circ}$.
Therefore,$\triangle AOF \sim \triangle BOD$ ($AA$ similarity).
So,$\frac{OA}{OB} = \frac{FA}{DB}$ .........$(1)$
Also,in $\triangle FAC$ and $\triangle EBC$:
$\angle A = \angle B = 90^{\circ}$ and $\angle FCA = \angle ECB$ (Vertically opposite angles).
Therefore,$\triangle FAC \sim \triangle EBC$ ($AA$ similarity).
So,$\frac{FA}{EB} = \frac{AC}{BC}$.
Since $OB$ is the perpendicular bisector of $DE$,$EB = DB$.
So,$\frac{FA}{DB} = \frac{AC}{BC}$ ......$(2)$
From $(1)$ and $(2)$,we have:
$\frac{AC}{BC} = \frac{OA}{OB}$.
Since $AC = OC - OA$ and $BC = OB - OC$,we get:
$\frac{OC - OA}{OB - OC} = \frac{OA}{OB}$.
$OB(OC - OA) = OA(OB - OC)$.
$OB \cdot OC - OA \cdot OB = OA \cdot OB - OA \cdot OC$.
$OB \cdot OC + OA \cdot OC = 2(OA \cdot OB)$.
Dividing both sides by $(OA \cdot OB \cdot OC)$:
$\frac{OB \cdot OC}{OA \cdot OB \cdot OC} + \frac{OA \cdot OC}{OA \cdot OB \cdot OC} = \frac{2(OA \cdot OB)}{OA \cdot OB \cdot OC}$.
$\frac{1}{OA} + \frac{1}{OB} = \frac{2}{OC}$.