Question

In the figure, $AD$ is a median of a triangle $ABC$ and $AM\perp BC$.

$A{D}^2-BC.DM+{\left[\frac{BC}{2}\right]}^2=$

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

Answer 1

Answer: (B)

$A{B}^2$

From the figure -------- given that
BD $=DC=\frac{1}{2}BCand\angle AMB={90}^{\circ }$
From the figure we have
${AM}^2+{BM}^2={AB}^2$
or ${AM}^2+{(BD-MD)}^2={AB}^2$
or $({AD}^2-{MD}^2)+{(BD-MD)}^2={AB}^2$
or ${AD}^2-{MD}^2+{BD}^2+{MD}^2-2BD\times MD={AB}^2$
or ${AD}^2+{\left(\frac{1}{2}BC\right)}^2-2\times \frac{BC}{2}\times MD={AB}^2(BD=\frac{1}{2}BC)$
or ${AD}^2-BC\times MD+{\left(\frac{1}{2}BC\right)}^2={AB}^2$