Question

In $\Delta ABC,AD$ is the median, then $A{B}^2+A{C}^2$ is _______.

• A. $A{D}^2+B{D}^2$
• B. $2(A{D}^2+B{D}^2)$
• C. $\frac{1}{2}(A{D}^2+B{D}^2)$
• D. $A{D}^2+2B{D}^2$

Answer 1

The correct option is B $2(A{D}^2+B{D}^2)$

$AD$ is the median of the $△ABC$.

Draw a perpendicular from the vertex $A$ to the side $BC$

In $△AEBand△AEC,$ using pythagoras theorem,

${AB}^2+{AC}^2={BE}^2+{AE}^2+{EC}^2+{AE}^2=2{AE}^2+{BE}^2+{EC}^2........\left(i\right)$

Now, ${BE}^2={(BD-ED)}^2{EC}^2={(ED+DC)}^2\therefore {AB}^2+{AC}^2=2{AE}^2+{(BD-ED)}^2+{(ED+DC)}^2=2{AE}^2+2{ED}^2+{BD}^2+{CD}^2-2BD.ED+2ED.CD$

But we have,

$BD=DC\left(asADisthemedian\right)\therefore BD.ED=ED.CD\therefore {AB}^2+{AC}^2=2{AE}^2+2{ED}^2+{BD}^2+{CD}^2=2{AE}^2+2{ED}^2+2{BD}^2=2({AE}^2+{ED}^2+{BD}^2)=2({AD}^2+{BD}^2)$

$(\because in△AED,{AE}^2+{ED}^2={AD}^2(usingpythagorastheorem\left)\right)$