Question

Question 8 (i)
In the given figure, O is a point in the interior of a triangle ABC, OD \(\perp\) BC, OE \(\perp\) AC and OF \(\perp\) AB.
Show that \((i) OA^2 + OB^2 + OC^2 - OD^2 - OE^2 - OF^2 = AF^2 + BD^2 + CE^2\)

Answer 1

(i)Construction:
Join OA, OB and OC.

Applying Pythagoras theorem in $\Delta AOF$, we have
$O{A}^2=O{F}^2+A{F}^2...\left(i\right)$
Similarly, in $\Delta BOD$
$O{B}^2=O{D}^2+B{D}^2...\left(ii\right)$
Similarly, in $\Delta COE$
$O{C}^2=O{E}^2+E{C}^2...\left(iii\right)$
Adding the above equations, we get,
$O{A}^2+O{B}^2+O{C}^2=O{F}^2+A{F}^2+O{D}^2+B{D}^2+O{E}^2+E{C}^2$
$O{A}^2+O{B}^2+O{C}^2-O{D}^2-O{E}^2-O{F}^2=A{F}^2+B{D}^2+C{E}^2$.