Q8 (2)   In Fig. 6.54, O is a point in the interior of a triangle ABC, OD $\perp$ BC, OE $\perp$  AC  and OF$\perp$ AB.              $AF^2 + BD^2 + CE^2 = AE^2 + CD^2 + BF^2.$

S seema garhwal

Join AO, BO, CO

In $\triangle$ AOF, by Pythagoras theorem,

$OA^2=OF^2+AF^2..................1$

In $\triangle$ BOD, by Pythagoras theorem,

$OB^2=OD^2+BD^2..................2$

In $\triangle$ COE, by Pythagoras theorem,

$OC^2=OE^2+EC^2..................3$

$OA^2+OB^2+OC^2=OF^2+AF^2+OD^2+BD^2+OE^2+EC^2$$\Rightarrow OA^2+OB^2+OC^2-OD^2-OE^2-OF^2=AF^2+BD^2+EC^2....................4$
$\Rightarrow (OA^2-OE^2)+(OC^2-OD^2)+(OB^2-OF^2)=AF^2+BD^2+EC^2$$\Rightarrow AE^2+CD^2+BF^2=AF^2+BD^2+EC^2$