Here we will prove that if two angles of a triangle are unequal, the greater angle has the greater side opposite to it.
Given: In ∆XYZ, ∠XYZ > ∠XZY
To Prove: XZ > XY
1. Let us assume that XZ is not greater than XY. Then XZ must be either equal to or less than XY.
Let XZ = XY
⟹ ∠XYZ = ∠XZY
2. But ∠XYZ ≠ ∠XZY
3. Again, XZ < XY
⟹ ∠XYZ < ∠XZY
4. But ∠XYZ is not less than ∠XZY.
5. Therefore, XZ is neither equal to nor less than XY.
6. Therefore, XZ > XY. (Proved)
1. Angles opposite to equal sides are equal.
2. Given, ∠XYZ > ∠XZY
3. The greater side of a triangle has the greater angle opposite to it.
4. Given that ∠XYZ > ∠XZY.
5. Both the assumptions are leading to contradictions.
6. From statement 5.
Note: The side opposite to the greatest angle in a triangle is the longest.