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. |
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