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 △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>2AL.
- 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 △ALB and △ELC, we have ∠ALB=∠ELC[Vertically opposite angles]AL=LE[By construction]BL=LC[AL is the median.]∴ △ALB≅△ELC[By SAS criterion] As corresponding parts of congruent triangles are equal, we haveAB=EC …(1)
- 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>AE⟹AC+AB>AE[From (1)]⟹AC+AB>2AL[∵ AE = 2AL] - 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)
- 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)
- Thus, the perimeter of a triangle is greater than the sum of its three medians.