Question

In given figure from a point O in the interior of a $△ABC$, perpendiculars OD, OE and OF are drawn to the sides BC, CA and AB respectively. Prove that :

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

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

Answer 1

Let O be a point in the interior of $△ABC$, and let $OD\perp BC,OE\perp CA$ and $OF\perp AB$

(i) In right triangles $△OFA,△ODB$ and $△OEC$ we have

${OA}^2={AF}^2+{OF}^2$

${OB}^2={BD}^2+{OD}^2$

and, ${OC}^2={CE}^2+{OE}^2$

Adding all these results, we get

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

$\Rightarrow$ ${AF}^2+{BD}^2+{CE}^2={OA}^2+{OB}^2+{OC}^2-{OD}^2-{OE}^2-{OF}^2$ $\left[Henceproved\right]$

(ii) In right triangles $△ODB$ and $△ODC$, we have

${OB}^2={OD}^2+{BD}^2$

and, ${OC}^2={OD}^2+{CD}^2$

$\therefore$ ${OB}^2-{OC}^2=({OD}^2+{BD}^2)-({OD}^2+{CD}^2)$

$\Rightarrow$ ${OB}^2-{OC}^2={BD}^2-{CD}^2$.......(i)

Similarly, we have

$\Rightarrow$ ${OC}^2-{OA}^2={CE}^2-{AE}^2$.......(ii)

and, ${OA}^2-{OB}^2={AF}^2-{BF}^2$.......(iii)

Adding (i), (ii) and (iii), we get

$({OB}^2-{OC}^2)+({OC}^2-{OA}^2)+({OA}^2-{OB}^2)=({BD}^2-{CD}^2)+({CE}^2-{AE}^2)+({AF}^2-{BF}^2)$

$\Rightarrow$ $({BD}^2+{CE}^2+{AF}^2)-({AE}^2+{CD}^2+{BF}^2)=0$

$\Rightarrow$ ${AF}^2+{BD}^2+{CE}^2={AE}^2+{BF}^2+{CD}^2$ $\left[Henceproved\right]$