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.

**Proof:**

1. In ∆XYP, ∠XYP = ∠XPY 2. ∠XPY = ∠XZY + ∠PYZ |
1. XY = XP 2. In ∆YPZ, exterior ∠XPY = Sum of interior opposite angles, ∠PZY |

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. |

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