Show that the perimeter of a triangle is greater than the sum of its three medians.


Answer:


Step by Step Explanation:
  1. Let AL,BM and CN be the medians of ABC.
      A B C N M L
  2. 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>2AL.
  3. We know that the median from a vertex to the opposite side of a triangle bisects the opposite side.
    Thus, we have BL=LC.
      A B C L
  4. Let's extend AL to E such that AL=LE and join the point E to the point C.
      A B C LE
  5. In ALB and ELC, we have ALB=ELC[Vertically opposite angles]AL=LE[By construction]BL=LC[AL is the median.] ALBELC[By SAS criterion] As corresponding parts of congruent triangles are equal, we haveAB=EC  (1)
  6. We know that the sum of any two sides of a triangle is greater than the third side.

    So, in AEC, we have AC+EC>AEAC+AB>AE[From (1)]AC+AB>2AL[ AE = 2AL]
  7. 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.  AB+AC>2AL(2)AB+BC>2BM(3)BC+AC>2CN(4)
  8. Adding (2), (3), and (4) we get (AB+AC+AB+BC+BC+AC)>(2AL+2BM+2CN)2(AB+BC+CA)>2(AL+BM+CN)(AB+BC+CA)>(AL+BM+CN)
  9. Thus, the perimeter of a triangle is greater than the sum of its three medians.

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