Question

In the above 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) $O{A}^2+O{B}^2+O{C}^2-O{D}^2-O{E}^2-O{F}^2=A{F}^2+B{D}^2+C{E}^2$
(ii) $A{F}^2+B{D}^2+C{E}^2=A{E}^2+C{D}^2+B{F}^2$

Answer 1

Construction: Join $AO$, $BO$ and $OC.$
(i)
In $△AOE,$
By Pythagoras Theorem,
$A{O}^2=A{E}^2+O{E}^2.....\left(1\right)$

In $△AOF,$
By Pythagoras Theorem,
$A{O}^2=A{F}^2+F{O}^2.....\left(2\right)$

In $△FBO,$
By Pythagoras Theorem,
$B{O}^2=B{F}^2+F{O}^2.....\left(3\right)$

In $△BDO,$
By Pythagoras Theorem,
$B{O}^2=B{D}^2+O{D}^2.....\left(4\right)$

In $△DOC,$
By Pythagoras Theorem,
$O{C}^2=O{D}^2+D{C}^2.....\left(5\right)$

In $△OCE,$
By Pythagoras Theorem,
$O{C}^2=O{E}^2+E{C}^2.....\left(6\right)$

In $△ABC,$
By Pythagoras Theorem,
$A{C}^2=A{B}^2+B{C}^2.....\left(7\right)$

Add equations $\left(2\right),\left(4\right)$ and $\left(6\right)$ then,
$A{O}^2+B{O}^2+O{C}^2=A{F}^2+F{O}^2+B{D}^2+O{D}^2+O{E}^2+E{C}^2$

$\Rightarrow A{O}^2+B{O}^2+O{C}^2-O{D}^2-O{E}^2-O{F}^2=A{F}^2+B{D}^2+E{C}^2$

(ii)

Subtract equation $\left(1\right)$ from $\left(2\right),$ we get,
$0=A{F}^2+F{O}^2-A{E}^2-O{E}^2$
$\Rightarrow A{F}^2+F{O}^2=A{E}^2+O{E}^2......\left(8\right)$

Subtract equation $\left(3\right)$ from $\left(4\right),$ we get,
$0=B{D}^2+O{D}^2-B{F}^2-F{O}^2$
$\Rightarrow B{D}^2+O{D}^2=B{F}^2+F{O}^2......\left(9\right)$

Subtract equation $\left(5\right)$ from $\left(6\right),$ we get,
$0=O{E}^2+E{C}^2-O{D}^2-D{C}^2$
$\Rightarrow O{E}^2+E{C}^2=O{D}^2+D{C}^2........\left(10\right)$

Add equations $\left(8\right),\left(9\right)$ and $\left(10\right)$ we get,
$A{F}^2+B{D}^2+C{E}^2=A{E}^2+C{D}^2+B{F}^2$

Hence, proved.