Question

From a point O in the interior of a $\Delta ABC$, perpendiculars OD, OE, & OF are drawn to the sides BC, CA & AB respectively. Prove that :
[i] $A{F}^2+B{D}^2+C{E}^2=O{A}^2+O{B}^2+O{C}^2-O{D}^2-O{E}^2-O{F}^2$
[ii] $A{F}^2+B{D}^2+C{E}^2=A{E}^2+C{D}^2+B{F}^2$

Answer 1

In right angled triangle $OFA$, we have

${OA}^2={AF}^2+{OF}^2$ ..........$\left(1\right)$

Using Pythagoras theorem,

In right angled triangle $OBD,$ we have

${OB}^2={OD}^2+{BD}^2$ ..........$\left(2\right)$

In right angled triangle $OEC,$ we have

${OC}^2={OE}^2+{CE}^2$ ..........$\left(3\right)$

Adding equations $\left(1\right),\left(2\right)$ and $\left(3\right)$ we get

${OA}^2+{OB}^2+{OC}^2={AF}^2+{OF}^2+{OD}^2+{BD}^2+{OE}^2+{CE}^2$

${OA}^2+{OB}^2+{OC}^2={AF}^2+{BD}^2+{CE}^2+{OF}^2+{OD}^2+{OE}^2$

${AF}^2+{BD}^2+{CE}^2={OA}^2+{OB}^2+{OC}^2-{OD}^2-{OE}^2-{OF}^2$ is proved.

$\left(ii\right)$ We can re-write the above proved result as

${AF}^2+{BD}^2+{CE}^2=({OA}^2-{OE}^2)+({OB}^2-{OF}^2)+({OC}^2-{OD}^2)$

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

Hence proved.