Question

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

Answer 1

Given : A $△$ABC in which AD is a median.


To prove : AB + AC > 2 AD
Construction :Produce AD to E such that
AD = DE. Join EC.

Proof : In $△$ADB and EDC, we have
AD = DE [By construction]
BD = DC [D is the mid point of BC]
and, $\angle ADB=\angle EDC$ [Vertically opp. angles]
So, by SAS criterion of congruence
$△ADB\cong △EDC\Rightarrow AB=EC$ [Corresponding parts of Congruent triangles are equal]
Now in $△$AEC, we have
AC + EC > AE [Sum of any two sides of a $△$ is greater than the third]
$\Rightarrow$ AC + AB > 2 AD
[AD = AE, i.e. AE = AD + DE = 2AD and EC = AB]