If `a+ b=2c,` then `(a)/( a-c)+ ( c) /(b-c)` is equal to

Options

A. 0
B. `-1`
C. `1`
D. `2`

Answer

To solve the problem, we need to evaluate the expression:

\[ \frac{a}{a-c} + \frac{c}{b-c} \]

given that \( a + b = 2c \).

### Step 1: Rewrite \( b \) in terms of \( a \) and \( c \)

From the equation \( a + b = 2c \), we can express \( b \) as:

\[ b = 2c - a \]

### Step 2: Substitute \( b \) into the expression

Now, we substitute \( b \) into the expression \( \frac{c}{b-c} \):

\[ b - c = (2c - a) - c = c - a \]

Thus, we can rewrite the expression as:

\[ \frac{a}{a-c} + \frac{c}{c-a} \]

### Step 3: Find a common denominator

The common denominator for the two fractions is \((a-c)(c-a)\). We can rewrite the expression as:

\[ \frac{a(c-a) + c(a-c)}{(a-c)(c-a)} \]

### Step 4: Simplify the numerator

Now, let's simplify the numerator:

\[ a(c-a) + c(a-c) = ac - a^2 + ac - c^2 = 2ac - a^2 - c^2 \]

### Step 5: Factor the numerator

Notice that we can factor the numerator:

\[ 2ac - a^2 - c^2 = -(a^2 - 2ac + c^2) = -(a-c)^2 \]

### Step 6: Substitute back into the expression

Now we substitute back into our expression:

\[ \frac{-(a-c)^2}{(a-c)(c-a)} \]

### Step 7: Simplify the expression

Since \( c-a = -(a-c) \), we can rewrite the denominator:

\[ (c-a) = -(a-c) \]

Thus, the expression simplifies to:

\[ \frac{-(a-c)^2}{(a-c)(-(a-c))} = \frac{-(a-c)^2}{-(a-c)^2} = 1 \]

### Conclusion

Therefore, the value of the expression \( \frac{a}{a-c} + \frac{c}{b-c} \) is:

\[ \boxed{1} \]

---