If x, y, z are three distinct complex numbers and a, b, c are three + ive real numbers satisfying the relation
`(a)/( | y -z|) = (b)/(| z - x|) = (c)/( | x - y|) ` then
`(a^(2))/( y - z) + (b^(2))/( z - x) + (c^(2))/( x - y) = 0 `

Answer

To prove the given statement, we start with the relation provided:

\[ \frac{a}{|y - z|} = \frac{b}{|z - x|} = \frac{c}{|x - y|} \]

Let's denote this common ratio by \( k \). Thus, we can express \( a, b, c \) in terms of \( k \):

\[ a = k |y - z|, \quad b = k |z - x|, \quad c = k |x - y| \]

Next, we need to compute \( a^2, b^2, c^2 \):

\[ a^2 = k^2 |y - z|^2, \quad b^2 = k^2 |z - x|^2, \quad c^2 = k^2 |x - y|^2 \]

Now, substituting these into the expression we want to prove:

\[ \frac{a^2}{y - z} + \frac{b^2}{z - x} + \frac{c^2}{x - y} \]

Substituting the values of \( a^2, b^2, c^2 \):

\[ = \frac{k^2 |y - z|^2}{y - z} + \frac{k^2 |z - x|^2}{z - x} + \frac{k^2 |x - y|^2}{x - y} \]

This simplifies to:

\[ = k^2 \left( |y - z|^2 \cdot \frac{1}{y - z} + |z - x|^2 \cdot \frac{1}{z - x} + |x - y|^2 \cdot \frac{1}{x - y} \right) \]

Using the property \( |z|^2 = z \cdot \overline{z} \), we can express \( |y - z|^2 \) as \( (y - z)(\overline{y - z}) \). Thus:

\[ = k^2 \left( (y - z) \overline{(y - z)} \cdot \frac{1}{y - z} + (z - x) \overline{(z - x)} \cdot \frac{1}{z - x} + (x - y) \overline{(x - y)} \cdot \frac{1}{x - y} \right) \]

This results in:

\[ = k^2 \left( \overline{(y - z)} + \overline{(z - x)} + \overline{(x - y)} \right) \]

Now, notice that the terms \( \overline{(y - z)} + \overline{(z - x)} + \overline{(x - y)} \) can be rearranged and simplified. Since \( x, y, z \) are distinct complex numbers, the sum of these terms will equal zero:

\[ \overline{(y - z)} + \overline{(z - x)} + \overline{(x - y)} = 0 \]

Thus, we conclude:

\[ \frac{a^2}{y - z} + \frac{b^2}{z - x} + \frac{c^2}{x - y} = k^2 \cdot 0 = 0 \]

This proves the required statement.