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

**Proof:**

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.

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