Consider a triangle ABC, the respective middles A′, B′, C′ of sides BC, CA, AB, and center of gravity G. (a)Show that AA′ +BB′ +CC′ = 3/2 (GA+GB +GC). (b)Show the ınequality GA + GB + GC > BC +CA+AB/2 . Deduce the inequality: AA′ + BB′ + CC′ > 3/4 (BC + CA+AB). (c) Let N be the symmetric of point A with respect to point A′. What order relationship exists in triangle ABN between sides AN and the sum of the other two sides? Deduce the inequality: AA′ < AB + AC /2 , then the inequality: AA′ + BB′ + CC′ < BC +CA+AB.