If ABC is a right triangle right-angled at B and M,
Question:

If $\mathrm{ABC}$ is a right triangle right-angled at $\mathrm{B}$ and $\mathrm{M}, \mathrm{N}$ are the mid-points of $\mathrm{AB}$ and $\mathrm{BC}$ respectively, then $4\left(\mathrm{AN}^{2}+\mathrm{CM}^{2}\right)=$

(a) $4 \mathrm{AC}^{2}$

(b) $5 \mathrm{AC}^{2}$

(c) $54 \mathrm{AC} 2$

(d) $6 \mathrm{AC}^{2}$

Solution:

M is the mid-point of AB.

$\therefore B M=A B 2$

N is the mid-point of BC.

$\therefore B N=B C 2$

Now,

$A N^{2}+C M^{2}=\left(A B^{2}+\left(\frac{1}{2} B C\right)^{2}\right)+\left(\left(\frac{1}{2} A B\right)^{2}+B C^{2}\right)$

$=A B^{2}+\frac{1}{4} B C^{2}+\frac{1}{4} A B^{2}+B C^{2}$

$=\frac{5}{4}\left(A B^{2}+B C^{2}\right)$

$\Rightarrow 4\left(A N^{2}+C M^{2}\right)=5 A C^{2}$

Hence option (b) is correct.