Question

prove that the sum of any two sides of a triangle is greater than twice the median with respect to the third side​

Answer 1

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

1. In ∆XYP and ∆ZPQ,

(i) YP = PZ.

(ii) XP = PQ

(iii) ∠XPY = ∠ZPQ

reason

(i) XP bisects YZ.

(ii) By construction.

(iii) Vertically opposite angles.

2. XYP ≅ ∆ZPQ

( reason: By SAS criterion of congruency.)

3. XY = ZQ. ( reason : CPCTC )

4. 4. In ∆XZQ, (XZ + ZQ) > XQ.

(reason : Sum of the two sides of a triangle is greater than the third side)

5. (XZ + XY) > (XP + PQ) ( reason : XY = ZQ, from statement 3 )

6. (XY + XZ) > 2 XP. ( reason : XP = PQ.)

hence proved