Prove that the sum of any two sides of a triangle is greater than twice the median drawn to the third side.


Answer:


Step by Step Explanation:
  1. Let AD be the median to the third side of the triangle ABC.

    We need to prove that AB+AC>2AD.
  2. We know that the median from a vertex to the opposite side of a triangle bisects the opposite side.
    Thus, we have BD=DC.
      A B C D
  3. Let's extend AD to E such that AD=DE and join the point E to the point C.
      A B C DE
  4. In ADB and EDC, we have ADB=EDC[Vertically opposite angles]AD=DE[By construction]BD=DC[AD is the median.] As corresponding parts of congruent triangles are equal, we haveAB = EC \space \space \ldots (1)
  5. 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 > 2AD && [\because \space \text{AE = 2AD}] \end{aligned}
  6. Thus, the sum of any two sides of a triangle is greater than twice the median drawn to the third side.

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