Here we will prove that if two sides of a triangle are unequal, the greater side has the greater angle opposite to it.
Given: In ∆XYZ, XZ > XY
To prove: ∠XYZ > ∠XZY.
Construction: From XZ, cut off XP such that XP equals XY. Join Y and P.
1. In ∆XYP, ∠XYP = ∠XPY
2. ∠XPY = ∠XZY + ∠PYZ
1. XY = XP
2. In ∆YPZ, exterior ∠XPY = Sum of interior opposite angles, ∠PZY (=∠XZY) and ∠PYZ.
3. Therefore, ∠XPY > ∠XZY.
4. Therefore, ∠XYP > ∠XZY.
5. But ∠XYZ > ∠ XYP.
6. Therefore, ∠XYZ > ∠XZY. (Proved)
3. From statement 2.
4. Using statements 1 in 3.
5. ∠XYZ = ∠XYP + ∠PYZ
6. Using statements 5 and 4.
Note: The angle opposite to the greater side in a triangle is the greatest in measure.
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