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

Explanation

Given in `Delta ABC` ,AD is a median

Constrution
Produce AD to a point E such that AD=DE and join CE
TP prove ` AC+AB gt 2 AD `
Proof in `Delta ABD and Delta ECD `
AD =DE [by construction ]
BD =CD [given AD is the median ]
`anngle ADB = angle CDE` [vertically opposite angle ]
`therefore Delta ABD cong Delta ECD ` [by SAS congruence rule ]
`rArr AB= CE ` [by CPCT ]
Now in `Delta AEC ,`
`AC+EC gt AE ` [Sum of two sides of a triagle is grater than the third side ]
`therefore AC+AB gt 2AD ` [from Eq. (i) and also taken that AD=AE ]