Italy
Show that the perimeter of a triangle is greater than the sum of its three medians.
Answer:
- Let ^@AL, BM^@ and ^@CN^@ be the medians of ^@ \triangle ABC. ^@
- We will first prove that the sum of any two sides of a triangle is greater than twice the median drawn to the third side, i.e, ^@AB + AC > 2 AL.^@
- We know that the median from a vertex to the opposite side of a triangle bisects the opposite side.
Thus, we have ^@BL = LC^@. - Let's extend ^@AL^@ to ^@E^@ such that ^@AL = LE^@ and join the point ^@E^@ to the point ^@C.^@
- In ^@\triangle ALB^@ and ^@\triangle ELC,^@ we have @^ \begin{aligned} & \angle ALB = \angle ELC && [\text{Vertically opposite angles}] \\ & AL = LE && [\text{By construction}] \\ & BL = LC && [\text{AL is the median.}] \\ \therefore \space & \triangle ALB \cong \triangle ELC && [\text{By SAS criterion}] \end{aligned}@^ As corresponding parts of congruent triangles are equal, we have@^AB = EC \space \space \ldots (1)@^
- We know that the sum of any two sides of a triangle is greater than the third side.
So, in ^@\triangle AEC, ^@ we have @^\begin{aligned} & AC + EC > AE \\ \implies & AC + AB > AE && [\text{From (1)}] \\ \implies & AC + AB > 2AL && [\because \space \text{AE = 2AL}] \end{aligned}@^ - From the above steps, we conclude that the sum of any two sides of a triangle is greater than twice the median drawn to the third side. @^ \begin{aligned}\therefore \space & AB + AC > 2AL && \ldots (2)\\ & AB + BC > 2BM && \ldots (3) \\ & BC + AC > 2CN && \ldots (4) \end{aligned}@^
- Adding (2), (3), and (4) we get @^\begin{aligned} & (AB + AC + AB + BC + BC + AC) > (2AL + 2BM + 2CN) \\ \implies & 2(AB + BC + CA) > 2(AL + BM + CN) \\ \implies & (AB + BC + CA) > (AL + BM + CN) \end{aligned}@^
- Thus, the perimeter of a triangle is greater than the sum of its three medians.
