Prove that the sum of any two sides of a triangle is greater than twice the median drawn to the third side.
Answer:
- Let AD be the median to the third side of the triangle
ABC.
We need to prove that AB+AC>2AD. - We know that the median from a vertex to the opposite side of a triangle bisects the opposite side.
Thus, we have BD=DC. - Let's extend AD to E such that AD=DE and join the point E to the point C.
- 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)
- 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} - Thus, the sum of any two sides of a triangle is greater than twice the median drawn to the third side.