According to the question,
We have, Δ ABC with median AD.
To prove:
AB + AC > 2AD
AB + BC > 2AD
BC + AC > 2AD
Construction:
Extend AD to E such that DE = AD
Join EC.
Proof:
In Δ ADB and Δ EDC ,
AD = ED [By construction]
∠1 = ∠2 [Vertically opposite angles are equal]
DB = DC [Given]
So, by SAS criterion of congruence, we have
Δ ADB ≅ Δ EDC
AB = EC [CPCT]
And ∠3 = ∠4 [CPCT]
Now, in Δ AEC,
Since sum of the lengths of any two sides of a triangle must be greater than the third side,
We have
AC + CE > AE
⇒ AC + CE > AD + DE
⇒ AC +CE > AD+ AD [AD = DE]
⇒ AC + CE > 2AD
⇒ AC + AB > 2AD [∵AB = CE]
Similarly,
We get,
AB + BC > 2AD and BC + AC >2AD.
Hence, proved.
