Question

In Fig., $O$ 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

Given: $OD\perp BC$, $OE\perp AC$ and $OF\perp AB$

Construction: Join point O with vertices of triangle A, B and C to get segments OA, OB and OC.

1) In right angled triangle $△AFO$, by Pythagoras theorem,

${AF}^2+{FO}^2={AO}^2$

$\therefore {AF}^2={AO}^2-{FO}^2$ (1)

In right angled triangle $△BDO$, by Pythagoras theorem,

${BD}^2+{DO}^2={BO}^2$

$\therefore {BD}^2={BO}^2-{DO}^2$ (2)

In right angled triangle $△CEO$, by Pythagoras theorem,

${CE}^2+{EO}^2={CO}^2$

$\therefore {CE}^2={CO}^2-{EO}^2$ (3)

Adding equations (1), (2) and (3), we get,

${AF}^2+{BD}^2+{CE}^2={AO}^2-{FO}^2+{BO}^2-{DO}^2+{CO}^2-{EO}^2$

$\therefore {AF}^2+{BD}^2+{CE}^2={AO}^2+{BO}^2+{CO}^2-{FO}^2-{DO}^2-{EO}^2$

Hence proved.

2) in right angled triangle $△AEO$, by Pythagoras theorem,

${AE}^2+{EO}^2={AO}^2$

$\therefore {AE}^2={AO}^2-{EO}^2$ (4)

In right angled triangle $△CDO$, by Pythagoras theorem,

${CD}^2+{DO}^2={CO}^2$

$\therefore {CD}^2={CO}^2-{DO}^2$ (5)

In right angled triangle $△BFO$, by Pythagoras theorem,

${BF}^2+{FO}^2={BO}^2$

$\therefore {BF}^2={BO}^2-{FO}^2$ (6)

Adding equations (4), (5) and (6), we get,

${AE}^2+{CD}^2+{BF}^2={AO}^2-{EO}^2+{CO}^2-{DO}^2+{BO}^2-{FO}^2$

$\therefore {AE}^2+{CD}^2+{BF}^2=({AO}^2-{FO}^2)+({BO}^2-{DO}^2)+({CO}^2-{EO}^2)$

From equations (1), (2) and (3),

$\therefore {AE}^2+{CD}^2+{BF}^2={AF}^2+{BF}^2+C{D}^2$

Hence proved