Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Let the midpoints of the sides $AB$, $BC$, and $CA$ be $D(l,0,0)$, $E(0,m,0)$, and $F(0,0,n)$ respectively.

By the midpoint theorem, the length of the line segment joining the midpoints of two sides of a triangle is half the length of the third side.

Therefore, $DE=\frac{1}{2}AC\Rightarrow A{C}^2=4D{E}^2$

$EF=\frac{1}{2}AB\Rightarrow A{B}^2=4E{F}^2$

$FD=\frac{1}{2}BC\Rightarrow B{C}^2=4F{D}^2$

Using the distance formula, we calculate the squares of the distances between the midpoints:

$D{E}^2=(l-0{)}^2+(0-m{)}^2+(0-0{)}^2=l^2+m^2$

$E{F}^2=(0-0{)}^2+(m-0{)}^2+(0-n{)}^2=m^2+n^2$

$F{D}^2=(0-l{)}^2+(0-0{)}^2+(n-0{)}^2=n^2+l^2$

Substituting these into the equations for the sides of the triangle:

$A{C}^2=4(l^2+m^2)$

$A{B}^2=4(m^2+n^2)$

$B{C}^2=4(n^2+l^2)$

Adding the squares of the sides of the triangle:

$(AB{)}^2+(BC{)}^2+(CA{)}^2=4(m^2+n^2)+4(n^2+l^2)+4(l^2+m^2)$

$(AB{)}^2+(BC{)}^2+(CA{)}^2=8(l^2+m^2+n^2)$

Thus, the required ratio is:

$\frac{(AB{)}^2+(BC{)}^2+(CA{)}^2}{l^2+m^2+n^2}=\frac{8(l^2+m^2+n^2)}{l^2+m^2+n^2}=8$

Answer: $8$