Question

If $△ABC$ is right angled at $B$ and $M,$ $N$ are the midpoints of $AB$ and $BC$ respectively, then $4(A{N}^2+C{M}^2)=$

• A. $4A{C}^2$
• B. $5A{C}^2$
• C. $\frac{5}{4}A{C}^2$
• D. $6A{C}^2$

Answer 1

The correct option is B $5A{C}^2$

Given, $△ABC$, $M$ is mid point of $AB$ and $N$ is mid point of $BC.$
In $△ABN,$
$A{N}^2=A{B}^2+B{N}^2$ (Pythagoras Theorem)
$A{N}^2=A{B}^2+(\frac{BC}{2}{)}^2$ $....\left(1\right)$
In $△BMC,$
$M{C}^2=B{M}^2+B{C}^2$ (Pythagoras Theorem)
$M{C}^2=B{C}^2+(\frac{AB}{2}{)}^2$ $....\left(2\right)$
Add $\left(1\right)$ and $\left(2\right),$
$A{N}^2+M{C}^2=A{B}^2+(\frac{BC}{2}{)}^2+B{C}^2+(\frac{AB}{2}{)}^2$

$A{N}^2+M{C}^2=\frac{5}{4}A{B}^2+\frac{5}{4}B{C}^2$

$4(A{N}^2+M{C}^2)=5(A{B}^2+B{C}^2)$

$4(A{N}^2+M{C}^2)=5A{C}^2$ (Pythagoras Theorem in $△ABC$)