Question

In the given figure, $AD$ is the median of a triangle $ABC$ and $AM\perp BC$. Then, $A{D}^2+BC.DM+{\left[\frac{BC}{2}\right]}^2=$

• A. $A{C}^2$
• B. $A{B}^2$
• C. $B{C}^2$
• D. none of the above

Answer 1

The correct option is A $A{C}^2$

Given CD $=BD=\frac{1}{2}BCand\angle AMB={90}^{\circ }$.
From the figure we get ${AC}^2={CM}^2+{AM}^2$
$={(CD+DM)}^2+({AD}^2-{DM}^2)$
$={CD}^2+{DM}^2+2CD\times DM+{AD}^2-{DM}^2$
$={AD}^2+{\left(\frac{1}{2}BC\right)}^2+(2\times \frac{1}{2}BC)DM(sinceCD=\frac{1}{2}BC)$
$={AD}^2+BC\times DM+{\left(\frac{1}{2}BC\right)}^2$