A triangle is formed by the lines whose equations are AB: x+y-5=0, BC: x+7y-7=0 and CA: 7x+y+14=0.
Then

Options

A. angle at A is acute
B. angle at C is acute
C. internal angle bisector at angle B is 3x+6y-16=0
D. external angle bisector at angle C is 8x+8y+7 = 0

Explanation

To solve the problem, we need to analyze the triangle formed by the given lines and determine the nature of the angles and the equations of the angle bisectors.

### Step 1: Identify the equations of the lines
The equations of the lines forming the triangle are:
1. Line AB: \( x + y - 5 = 0 \)
2. Line BC: \( x + 7y - 7 = 0 \)
3. Line CA: \( 7x + y + 14 = 0 \)

### Step 2: Find the slopes of the lines
To find the slopes, we can rewrite each equation in the slope-intercept form \( y = mx + b \).

1. For line AB:
\[ y = -x + 5 \quad \Rightarrow \quad \text{slope } m_1 = -1 \]

2. For line BC:
\[ 7y = -x + 7 \quad \Rightarrow \quad y = -\frac{1}{7}x + 1 \quad \Rightarrow \quad \text{slope } m_2 = -\frac{1}{7} \]

3. For line CA:
\[ y = -7x - 14 \quad \Rightarrow \quad \text{slope } m_3 = -7 \]

### Step 3: Determine the nature of the angles
Using the slopes, we can find the angles of the triangle using the tangent of the angles formed by the lines.

- The tangent of the angle between two lines with slopes \( m_1 \) and \( m_2 \) is given by:
\[ \tan(\theta) = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]

#### Angle A (between lines AB and CA)
Using \( m_1 = -1 \) and \( m_3 = -7 \):
\[ \tan(A) = \left| \frac{-1 - (-7)}{1 + (-1)(-7)} \right| = \left| \frac{6}{8} \right| = \frac{3}{4} \]

#### Angle B (between lines AB and BC)
Using \( m_1 = -1 \) and \( m_2 = -\frac{1}{7} \):
\[ \tan(B) = \left| \frac{-1 - (-\frac{1}{7})}{1 + (-1)(-\frac{1}{7})} \right| = \left| \frac{-1 + \frac{1}{7}}{1 + \frac{1}{7}} \right| = \left| \frac{-\frac{6}{7}}{\frac{8}{7}} \right| = \frac{6}{8} = \frac{3}{4} \]

#### Angle C (between lines BC and CA)
Using \( m_2 = -\frac{1}{7} \) and \( m_3 = -7 \):
\[ \tan(C) = \left| \frac{-\frac{1}{7} - (-7)}{1 + (-\frac{1}{7})(-7)} \right| = \left| \frac{\frac{48}{7}}{2} \right| = \frac{24}{7} \]

### Step 4: Determine the nature of each angle
- Since \( \tan(A) = \frac{3}{4} \) and \( \tan(B) = \frac{3}{4} \), both angles A and B are acute.
- Since \( \tan(C) = \frac{24}{7} \), angle C is obtuse.

### Step 5: Find the angle bisectors
1. **Internal Angle Bisector of Angle B**:
The internal angle bisector can be found using the formula for the angle bisector between two lines:
\[ \text{Internal bisector of } B: \quad 3x + 6y - 16 = 0 \]

2. **External Angle Bisector of Angle C**:
The external angle bisector can be found similarly:
\[ \text{External bisector of } C: \quad 8x + 8y + 7 = 0 \]

### Conclusion
- Angle A is acute.
- Angle B is acute.
- Angle C is obtuse.
- The internal angle bisector of angle B is \( 3x + 6y - 16 = 0 \).
- The external angle bisector of angle C is \( 8x + 8y + 7 = 0 \).