Question

Prove that any two sides of a triangle are together
greater than the median dhawn to the third side​

Answer 1

Theorem: The sum of any two sides of a triangle is greater than twice the median drawn to the third side...

Assumption: let ABC be a triangle

Given: ΔABC in which AD is a median.

To Prove: AB+AC>2AD

Construction: Produce AD to E such that AD = DE. Join EC.

Proof: In ΔADB and ΔEDC,

AD = DE. (By construction)

BD = DC. (∵ D is midpoint of BC)

∠ADB = ∠EDC (Ver. Opp. Angles)

By SAS congruency,

Δ ADB ≅ ΔEDC

AB = EC. (CPCT)

In ΔAEC,

As the sum of two sides is greater than third side,

AC + EC > AE

∵ AD = DE

∴ AE = AD + DE and EC = AB

⇒ AC + AB > 2 AD

[Hence: Proved] Q.E.D