Greater Angle has the Greater Side Opposite to It

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

$Greater Angle has the Greater Side Opposite to It$

To Prove: XZ > XY

Proof:

Statement

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)

Reason

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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