# Greater Side has the Greater Angle Opposite to it

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:

 Statement1. In ∆XYP, ∠XYP = ∠XPY2. ∠XPY = ∠XZY + ∠PYZ Reason1. XY = XP2. 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 + ∠PYZ6. 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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