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.


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

SAS Criterion of Congruency

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

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

Sum Of Any Two Sides Is Greater Than Twice The Median




1. In ∆XYP and ∆ZPQ,

(i) YP = PZ.

(ii) XP = PQ

(iii) ∠XPY = ∠ZPQ


(i) XP bisects YZ.

(ii) By construction.

(iii) Vertically opposite angles.

 2. XYP ≅ ∆ZPQ

2. By SAS criterion of congruency.

3. XY = ZQ.


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.

9th Grade Math

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