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 $AF^{2} + BD^{2} + CE^{2} = AE^{2} + CD^{2} + BF^{2}$.

(N/A) Join $OA$,$OB$,and $OC$.
In right-angled triangles $\triangle AFO$,$\triangle BDO$,and $\triangle CEO$,by applying the Pythagoras theorem:
$OA^{2} = AF^{2} + OF^{2} \implies AF^{2} = OA^{2} - OF^{2}$ ... $(i)$
$OB^{2} = BD^{2} + OD^{2} \implies BD^{2} = OB^{2} - OD^{2}$ ... $(ii)$
$OC^{2} = CE^{2} + OE^{2} \implies CE^{2} = OC^{2} - OE^{2}$ ... $(iii)$
Adding $(i)$,$(ii)$,and $(iii)$,we get:
$AF^{2} + BD^{2} + CE^{2} = (OA^{2} - OF^{2}) + (OB^{2} - OD^{2}) + (OC^{2} - OE^{2})$ ... $(iv)$
Similarly,in right-angled triangles $\triangle AEO$,$\triangle CDO$,and $\triangle BFO$:
$OA^{2} = AE^{2} + OE^{2} \implies AE^{2} = OA^{2} - OE^{2}$
$OC^{2} = CD^{2} + OD^{2} \implies CD^{2} = OC^{2} - OD^{2}$
$OB^{2} = BF^{2} + OF^{2} \implies BF^{2} = OB^{2} - OF^{2}$
Adding these,we get:
$AE^{2} + CD^{2} + BF^{2} = (OA^{2} - OE^{2}) + (OC^{2} - OD^{2}) + (OB^{2} - OF^{2})$
Rearranging the terms,this is equal to the right-hand side of equation $(iv)$.
Thus,$AF^{2} + BD^{2} + CE^{2} = AE^{2} + CD^{2} + BF^{2}$.