Question

In any triangle $ABC$, $A{B}^2+A{C}^2=3(A{O}^2+O{C}^2)$.
where $O$ is mid-point of $BC$.

• A. True
• B. False

Answer 1

The correct option is B False

1) Refer to the figure. O is the midpoint of side $BC$ and $segAD\perp segBC$

2) Consider right angled $△ADB$. By Pythagoras theorem,

${AB}^2=A{D}^2+D{B}^2$ (1)

Similarly, in right angled $△ADC$, by Pythagoras theorem,

${AC}^2={AD}^2+{DC}^2$ (2)

Adding equation (1) and (2), we get,

${AB}^2+{AC}^2={AD}^2+{DB}^2+{AD}^2+{DC}^2$

$\therefore {AB}^2+{AC}^2=2{AD}^2+{DB}^2+{DC}^2$

From the figure, it is evident that,

$BD=BO+OD$ and $CD=CO-OD$

$\therefore {AB}^2+{AC}^2=2{AD}^2+{(BO+OD)}^2+{(CO-OD)}^2$

$\therefore {AB}^2+{AC}^2=2{AD}^2+{BO}^2+2BO.OD+{OD}^2+{CO}^2+2CO.OD+{OD}^2$

But, $BO=CO$ (GIven)

$\therefore {AB}^2+{AC}^2=2{AD}^2+{BO}^2+2BO.OD+{OD}^2+{BO}^2-2BO.OD+{OD}^2$

$\therefore {AB}^2+{AC}^2=2{AD}^2+2{BO}^2+2{OD}^2$

$\therefore {AB}^2+{AC}^2=2({AD}^2+{BO}^2+{OD}^2)$ (3)

Now, refer to the figure. In right angled $△AOD$ by Pythagoras theorem,

${AD}^2+{OD}^2={AO}^2$

Thus, from (3),

$\therefore {AB}^2+{AC}^2=2({AO}^2+{BO}^2)$

$\therefore {AB}^2+{AC}^2=2({AO}^2+O{C}^2)$