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 (=∠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.

**From ****Greater Side has the Greater Angle Opposite to It**** to HOME PAGE**

**Didn't find what you were looking for? Or want to know more information
about Math Only Math.
Use this Google Search to find what you need.**

## New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.