In figure, O is a point in the interior of a triangle ABC,

Solution:

(i) In right angled $\Delta \mathrm{OFA}$,



$\mathrm{OA}^{2}=\mathrm{OF}^{2}+\mathrm{AF}^{2}$ (Pythagoras theorem)

$\Rightarrow \quad \quad \mathrm{OA}^{2}-\mathrm{OF}^{2}=\mathrm{AF}^{2}$.(1)

Similarly, $\quad \mathrm{OB}^{2}-\mathrm{OD}^{2}=\mathrm{BD}^{2}$ ...(2)

and $\quad \mathrm{OC}^{2}-\mathrm{OE}^{2}=\mathrm{CE}^{2}$ ...(3)

Adding $(1),(2)$ and $(3)$, we get

$\mathrm{OA}^{2}+\mathrm{OB}^{2}+\mathrm{OC}^{2}-\mathrm{OD}^{2}-\mathrm{OE}^{2}-\mathrm{OF}^{2}$

$=\mathrm{AF}^{2}+\mathrm{BD}^{2}+\mathrm{CE}^{2}$

(ii) We have proved that

$\mathrm{OA}^{2}+\mathrm{OB}^{2}+\mathrm{OC}^{2}-\mathrm{OD}^{2}-\mathrm{OE}^{2}-\mathrm{OF}^{2}$

$=\mathrm{AF}^{2}+\mathrm{BD}^{2}+\mathrm{CE}^{2}$ ...(4)

Similarly, we can prove that

$\mathrm{OA}^{2}+\mathrm{OB}^{2}+\mathrm{OC}^{2}-\mathrm{OD}^{2}-\mathrm{OE}^{2}-\mathrm{OF}^{2}$

$=\mathrm{BF}^{2}+\mathrm{CD}^{2}+\mathrm{AE}^{2}$ ...(5)

From (4) and (5), we have

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