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 |
5. (XZ + XY) > (XP + PQ). |
5. XY = ZQ, from statement 3. |
6. (XY + XZ) > 2 XP. (Proved) |
6. XP = PQ. |
From Sum Of Any Two Sides Is Greater Than Twice The Median to HOME PAGE
Didn’t find what you were looking for? Or want to know more information
about Math Only Math.
Use this Google Search to find what you need.