Question

prove that in triangle the larger angle parallel be opposite to longest side ​

Answer 1

Here we will prove that if two sides of a triangle are unequal, the greater side has the greater angle opposite to it.

To prove: ∠XYZ > ∠XZY.

Construction: From XZ, cut off XP such that XP equals XY. Join Y and P.

Proof:

Statement

1. In ∆XYP, ∠XYP = ∠XPY

2. ∠XPY = ∠XZY + ∠PYZ

Reason

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.