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 $\left(i\right)O{A}^2+O{B}^2+O{C}^2-O{D}^2-O{E}^2-O{F}^2=A{F}^2+B{D}^2+C{E}^2$

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,
$O{A}^2+O{B}^2+O{C}^2=O{F}^2+A{F}^2+O{D}^2+B{D}^2+O{E}^2+E{C}^2$
$O{A}^2+O{B}^2+O{C}^2-O{D}^2-O{E}^2-O{F}^2=A{F}^2+B{D}^2+C{E}^2$.