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