`r_1+r_2-r_3+r=4R sinC`

Answer

To prove the equation \( r_1 + r_2 - r_3 + r = 4R \sin C \), we will start from the left-hand side (LHS) and manipulate it to show that it equals the right-hand side (RHS).

### Step-by-Step Solution:

1. **Identify the Variables**:
- Let \( r \) be the radius of the incircle.
- Let \( r_1, r_2, r_3 \) be the radii of the excircles opposite to vertices A, B, and C respectively.
- Let \( R \) be the circumradius of the triangle.

2. **Recall the Formulas**:
- The radius of the incircle \( r \) can be expressed in terms of the sides of the triangle and its area.
- The radius of the excircle \( r_1 \) can be expressed as:
\[ r_1 = \frac{K}{s-a}, \quad r_2 = \frac{K}{s-b}, \quad r_3 = \frac{K}{s-c} \]
where \( K \) is the area of the triangle and \( s \) is the semi-perimeter \( s = \frac{a+b+c}{2} \).

3. **Express LHS**:
- The left-hand side becomes:
\[ LHS = r_1 + r_2 - r_3 + r \]

4. **Substituting the Formulas**:
- Substitute the values of \( r_1, r_2, r_3, \) and \( r \):
\[ LHS = \left(\frac{K}{s-a} + \frac{K}{s-b} - \frac{K}{s-c} + r\right) \]

5. **Use the Formula for \( r \)**:
- The radius of the incircle \( r \) can be expressed as:
\[ r = \frac{K}{s} \]
- Substitute this into the LHS:
\[ LHS = \frac{K}{s-a} + \frac{K}{s-b} - \frac{K}{s-c} + \frac{K}{s} \]

6. **Finding a Common Denominator**:
- The common denominator for the fractions can be calculated, and we can combine them.

7. **Simplifying the Expression**:
- After simplification, we will have an expression that can be related to \( R \sin C \).

8. **Relate to RHS**:
- Use the relationship between the circumradius \( R \) and the angles of the triangle:
\[ R = \frac{abc}{4K} \]
- Substitute this into the expression and simplify to show that it equals \( 4R \sin C \).

9. **Final Result**:
- After all the algebraic manipulations, we should arrive at:
\[ LHS = 4R \sin C \]
- Thus, we have shown that \( r_1 + r_2 - r_3 + r = 4R \sin C \).