Philippines
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 > 2 AD.^@ - 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 ^@\triangle ADB^@ and ^@\triangle EDC,^@ we have @^ \begin{aligned} & \angle ADB = \angle EDC && [\text{Vertically opposite angles}] \\ & AD = DE && [\text{By construction}] \\ & BD = DC && [\text{AD is the median.}] \\ \therefore \space & \triangle ADB \cong \triangle EDC && [\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 > 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.
