Question

If AD, BE and CF are the medians of an equilateral triangle ABC, then which of the following statements is valid?

• A. AB² + BC² + AC² = AD² + BE² + CF²
• B. 2 (AB² + BC² + AC²) = 3 (AD² + BE² + CF²)
• C. 3 (AB² + BC² + AC²) = 4 (AD² + BE² + CF²)
• D. AB² + BC² + AC² = 3 (AD² + BE² + CF²)

Answer 1

The correct option is C

3 (AB² + BC² + AC²) = 4 (AD² + BE² + CF²)

For triangle ABD,
${AB}^2={AD}^2+{\left(\frac{1}{2}BC\right)}^2$
For triangle ACD,
${AC}^2={AD}^2+{\left(\frac{1}{2}BC\right)}^2$
Adding the two equations, we get
${AB}^2+{AC}^2=2{AD}^2+2{\left(\frac{1}{2}BC\right)}^2$

$\Rightarrow {AB}^2+{AC}^2=2{AD}^2+\frac{1}{2}{BC}^2$ (1)

Similarly,

${BC}^2+{AB}^2=2{BE}^2+\frac{1}{2}{AC}^2$ (2)

and ${BC}^2+{AC}^2=2{CF}^2+\frac{1}{2}{AB}^2$ (3)

Adding (1), (2) and (3), we get:

$2({AB}^2+{BC}^2+{AC}^2)=2({AD}^2+{BE}^2+{CF}^2)+\frac{1}{2}(A{B}^2+B{C}^2+A{C}^2)$

$\Rightarrow \frac{3}{2}(A{B}^2+B{C}^2+A{C}^2)=2(A{D}^2+B{E}^2+C{F}^2)$

$\Rightarrow 3(A{B}^2+B{C}^2+A{C}^2)=4(A{D}^2+B{E}^2+C{F}^2)$