Question

If AD, BE and CF are the medians of an equilateral triangle ABC, then the true statement is -

• 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)

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

$⟹$ ${AB}^2$ + ${AC}^2$ = 2 ${AD}^2$ + $\frac{1}{2}$${BC}^2$

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

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

Adding we get

3( ${AB}^2$ + ${BC}^2$ + ${AC}^2$) = 4 ( ${AD}^2$ + ${BE}^2$ + ${CF}^2$)