# Sum Of Any Two Sides Is Greater Than Twice The Median

Here we will prove that in a triangle the sum of any two sides is greater than twice the median which bisects the remaining side.

Solution:

Given: In ∆XYZ, XP is the median that bisects YZ at P.

To prove: (XY + XZ) > 2XP.

Construction: Produce XP to Q such that XP = PQ. Join Z and Q.

Proof:

 Statement Reason 1. In ∆XYP and ∆ZPQ,(i) YP = PZ.(ii) XP = PQ (iii) ∠XPY = ∠ZPQ 1. (i) XP bisects YZ.(ii) By construction. (iii) Vertically opposite angles. 2. XYP ≅ ∆ZPQ 2. By SAS criterion of congruency. 3. XY = ZQ. 3. CPCTC. 4. In ∆XZQ, (XZ + ZQ) > XQ. 4. Sum of the two sides of a triangle is greater than the third side. 5. (XZ + XY) > (XP + PQ). 5. XY = ZQ, from statement 3. 6. (XY + XZ) > 2 XP. (Proved) 6. XP = PQ.