Question : In a triangle ABC, AD, BE, and CF are three medians. The perimeter of ABC is:
Option 1: equal to $(\overline {AD}+\overline{BE}+\overline{CF})$
Option 2: greater than $(\overline {AD}+\overline{BE}+\overline{CF})$
Option 3: less than $(\overline {AD}+\overline{BE}+\overline{CF})$
Option 4: none of these
Correct Answer: greater than $(\overline {AD}+\overline{BE}+\overline{CF})$
Solution : We know that the sum of any two sides of a triangle is greater than twice the median bisecting the third side. Here, D, E, and F are the midpoints of the sides BC, AC, and AB respectively. ∴ AB + AC > 2AD --(1) AB + BC > 2BE --(2) BC + AC > 2CF --(3) Adding all three equations, we get 2(AB + BC + AC) > 2(AD + BE + CF) ⇒ AB + BC + AC > AD + BE + CF ⇒ AB + BC + AC > $(\overline {AD}+\overline{BE}+\overline{CF})$ Hence, the correct answer is greater than $(\overline {AD}+\overline{BE}+\overline{CF})$.
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