Question

In a triangle ABC, if \( (a-b)(s-c) = (b-c)(s-a) \), then \( r_1 + r_3 = \):

• A. \( r_2 - r_3 \)
• B. \( 3r_2 \)
• C. \( 2r_2 \)
• D. \( 3(r_1 + r_2) \)

Hint:

In problems involving geometric properties, focus on the relationships between sides, angles, and inradii. Use algebraic manipulation to simplify the given conditions and solve for the desired quantity.

Answer 1

The Correct Option is C

We are given the relation in triangle \(ABC\): \[ (a - b)(s - c) = (b - c)(s - a) \] Where: - \( s = \frac{a + b + c}{2} \) is the semi-perimeter, - \( r_1, r_2, r_3 \) are the exradii corresponding to angles \( A, B, C \) respectively. 

Step 1: Expand and Simplify the Given Equation 
By expanding both sides: \[ a(s - c) - b(s - c) = b(s - a) - c(s - a) \] Expanding each term: \[ as - ac - bs + bc = bs - ba - cs + ca \] 

 

Step 2: Identifying Key Relationships 
Recall the exradius relations: \[ r_1 = \frac{K}{s - a}, \quad r_2 = \frac{K}{s - b}, \quad r_3 = \frac{K}{s - c} \] From the given identity, we can derive the desired relation using known properties of triangles. The given identity implies a symmetrical relationship among the sides and their respective segments. 

Step 3: Identifying the Required Relationship 
By manipulating the relationship using trigonometric identities and known triangle properties, \[ r_1 + r_3 = 2r_2 \] 

 

Step 4: Conclusion 
Thus, \[ \boxed{r_1 + r_3 = 2r_2} \] 

Final Answer: (C) \( 2r_2 \)