Question

Question 10
Prove that sum of any two sides of a triangle is greater than twice the median with respect to the third side.

Answer 1

Given In $\Delta ABC,$ AD is a median.

Construction:
Produce AD is a point E such that AD = DE and join CE.
To prove: AC + AB > 2AD

Proof:
$In\Delta ABDand\Delta ECD$,
AD = DE [by construction]
BD = CD [given AD is the median]
$and\angle ADB=\angle CDE$ [vertically opposite angle]
$\therefore \Delta ABD=\Delta ECD$ [by SAS congrence rule]
$\Rightarrow$ EC=AB [By CPCT] .... (i)
Now, $In\Delta AEC$,
AC + EC > AE [sum of two sides of a triangle is greater than the third side]
$\therefore AC+AB>2AD$ [From Eq. (i) and also taken that AD = DE]
Hence proved.