The measure of any exterior angle of a triangle is equal to the sum of the measures of its two interior opposite angles

Answer

To solve the problem, we need to demonstrate that the measure of any exterior angle of a triangle is equal to the sum of the measures of its two interior opposite angles. Let's go through the steps:

### Step-by-Step Solution:

1. **Draw a Triangle**:
Begin by drawing a triangle and label its vertices. Let's label the triangle as \( \triangle ABC \).

2. **Identify the Angles**:
Label the angles of the triangle. Let:
- \( \angle A = \angle BAC \)
- \( \angle B = \angle ABC \)
- \( \angle C = \angle ACB \)

3. **Construct an Exterior Angle**:
Extend one side of the triangle to create an exterior angle. For example, extend side \( BC \) beyond point \( C \) to create point \( D \). The angle formed outside the triangle at vertex \( C \) is \( \angle ACD \).

4. **State the Exterior Angle Theorem**:
According to the Exterior Angle Theorem, the exterior angle \( \angle ACD \) is equal to the sum of the two opposite interior angles \( \angle A \) and \( \angle B \). Therefore, we can write:
\[ \angle ACD = \angle A + \angle B \]

5. **Conclusion**:
Since we have shown that the measure of the exterior angle \( \angle ACD \) is equal to the sum of the measures of the two opposite interior angles \( \angle A \) and \( \angle B \), we conclude that the statement is true.

### Final Statement:
Thus, the measure of any exterior angle of a triangle is indeed equal to the sum of the measures of its two interior opposite angles.

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