In the figure,$O$ is a point in the interior of a triangle $ABC$,$OD \perp BC$,$OE \perp AC$,and $OF \perp AB$. Show that $OA^{2} + OB^{2} + OC^{2} - OD^{2} - OE^{2} - OF^{2} = AF^{2} + BD^{2} + CE^{2}$.

(N/A) Join $OA$,$OB$,and $OC$.
Applying the Pythagoras theorem in $\triangle AOF$,we obtain:
$OA^{2} = OF^{2} + AF^{2}$ --- $(i)$
Similarly,in $\triangle BOD$:
$OB^{2} = OD^{2} + BD^{2}$ --- $(ii)$
Similarly,in $\triangle COE$:
$OC^{2} = OE^{2} + CE^{2}$ --- $(iii)$
Adding equations $(i)$,$(ii)$,and $(iii)$,we get:
$OA^{2} + OB^{2} + OC^{2} = OF^{2} + AF^{2} + OD^{2} + BD^{2} + OE^{2} + CE^{2}$
Rearranging the terms,we get:
$OA^{2} + OB^{2} + OC^{2} - OD^{2} - OE^{2} - OF^{2} = AF^{2} + BD^{2} + CE^{2}$
Hence,the result is proved.