If g, h, k denotes the side of a pedal triangle, then prove that
`(g)/(a^(2))+ (h)/(b^(2))+ (k)/(c^(2))=(a^(2)+b^(2) +c^(2))/(2 abc)`

Answer

To prove the equation

\[ \frac{g}{a^2} + \frac{h}{b^2} + \frac{k}{c^2} = \frac{a^2 + b^2 + c^2}{2abc} \]

where \(g\), \(h\), and \(k\) are the sides of a pedal triangle corresponding to the angles \(A\), \(B\), and \(C\) of triangle \(ABC\), we start by using the known relationships for the sides of the pedal triangle:

1. **Identify the sides of the pedal triangle**:
\[ g = a \cos A, \quad h = b \cos B, \quad k = c \cos C \]

2. **Substituting the values of \(g\), \(h\), and \(k\)** into the left-hand side of the equation:
\[ \frac{g}{a^2} + \frac{h}{b^2} + \frac{k}{c^2} = \frac{a \cos A}{a^2} + \frac{b \cos B}{b^2} + \frac{c \cos C}{c^2} \]
This simplifies to:
\[ \frac{\cos A}{a} + \frac{\cos B}{b} + \frac{\cos C}{c} \]

3. **Using the cosine rule**:
Recall the cosine rule:
\[ \cos A = \frac{b^2 + c^2 - a^2}{2bc}, \quad \cos B = \frac{c^2 + a^2 - b^2}{2ca}, \quad \cos C = \frac{a^2 + b^2 - c^2}{2ab} \]

4. **Substituting the cosine values**:
Substitute these expressions back into our equation:
\[ \frac{\frac{b^2 + c^2 - a^2}{2bc}}{a} + \frac{\frac{c^2 + a^2 - b^2}{2ca}}{b} + \frac{\frac{a^2 + b^2 - c^2}{2ab}}{c} \]

5. **Simplifying each term**:
This gives:
\[ \frac{b^2 + c^2 - a^2}{2abc} + \frac{c^2 + a^2 - b^2}{2abc} + \frac{a^2 + b^2 - c^2}{2abc} \]

6. **Combining the fractions**:
Combine the fractions:
\[ \frac{(b^2 + c^2 - a^2) + (c^2 + a^2 - b^2) + (a^2 + b^2 - c^2)}{2abc} \]

7. **Simplifying the numerator**:
The numerator simplifies to:
\[ b^2 + c^2 - a^2 + c^2 + a^2 - b^2 + a^2 + b^2 - c^2 = a^2 + b^2 + c^2 \]

8. **Final expression**:
Therefore, we have:
\[ \frac{a^2 + b^2 + c^2}{2abc} \]

9. **Conclusion**:
Thus, we have shown that:
\[ \frac{g}{a^2} + \frac{h}{b^2} + \frac{k}{c^2} = \frac{a^2 + b^2 + c^2}{2abc} \]

Hence, the equation is proved.