`(r_1(r_2+r_3))/a = (r_2(r_3+r_1))/b = (r_3(r_1+r_2))/c`

Answer

To solve the equation \((r_1(r_2+r_3))/a = (r_2(r_3+r_1))/b = (r_3(r_1+r_2))/c\), we will follow these steps:

### Step 1: Understand the Variables
We know that \(r_1\), \(r_2\), and \(r_3\) are the inradii of the triangle, and \(a\), \(b\), and \(c\) are the lengths of the sides opposite to the vertices where the inradii are located.

### Step 2: Use the Formula for Inradius
The inradius \(r\) of a triangle can be expressed as:
\[ r_1 = \frac{\Delta}{s - a}, \quad r_2 = \frac{\Delta}{s - b}, \quad r_3 = \frac{\Delta}{s - c} \]
where \(\Delta\) is the area of the triangle and \(s\) is the semi-perimeter given by \(s = \frac{a + b + c}{2}\).

### Step 3: Substitute the Inradii into the Equation
We will substitute \(r_1\), \(r_2\), and \(r_3\) into the equation:
1. For \(r_1\):
\[ \frac{r_1(r_2 + r_3)}{a} = \frac{\frac{\Delta}{s - a} \left(\frac{\Delta}{s - b} + \frac{\Delta}{s - c}\right)}{a} \]
2. For \(r_2\):
\[ \frac{r_2(r_3 + r_1)}{b} = \frac{\frac{\Delta}{s - b} \left(\frac{\Delta}{s - c} + \frac{\Delta}{s - a}\right)}{b} \]
3. For \(r_3\):
\[ \frac{r_3(r_1 + r_2)}{c} = \frac{\frac{\Delta}{s - c} \left(\frac{\Delta}{s - a} + \frac{\Delta}{s - b}\right)}{c} \]

### Step 4: Simplify Each Expression
Now we simplify each expression:
1. For the first expression:
\[ \frac{\Delta^2 \left(\frac{1}{s - b} + \frac{1}{s - c}\right)}{a(s - a)} \]
Taking LCM:
\[ = \frac{\Delta^2 \cdot (s - c + s - b)}{a(s - a)(s - b)(s - c)} = \frac{\Delta^2 \cdot (2s - (b + c))}{a(s - a)(s - b)(s - c)} \]

2. For the second expression:
\[ \frac{\Delta^2 \cdot (s - c + s - a)}{b(s - b)(s - c)} = \frac{\Delta^2 \cdot (2s - (c + a))}{b(s - a)(s - b)(s - c)} \]

3. For the third expression:
\[ \frac{\Delta^2 \cdot (s - a + s - b)}{c(s - c)(s - a)} = \frac{\Delta^2 \cdot (2s - (a + b))}{c(s - a)(s - b)(s - c)} \]

### Step 5: Set the Expressions Equal
Now we set the three expressions equal to each other:
\[ \frac{\Delta^2 \cdot (2s - (b + c))}{a(s - a)(s - b)(s - c)} = \frac{\Delta^2 \cdot (2s - (c + a))}{b(s - a)(s - b)(s - c)} = \frac{\Delta^2 \cdot (2s - (a + b))}{c(s - a)(s - b)(s - c)} \]

### Step 6: Conclude
Since \(\Delta^2\) and the common denominators are the same, we can conclude that:
\[ \frac{2s - (b + c)}{a} = \frac{2s - (c + a)}{b} = \frac{2s - (a + b)}{c} \]
This shows that the original equation holds true.