Question

1/b+c+1/c+a=2/a+b then prooved 2c^2=a^2+b^2​

Answer 1

Answer:

Here, 1/(b + c) + 1/(c + a) = 2/(a + b)

or, (c + a + b + c)/{(b + c)(c + a)} = 2/(a + b)

or, (c + a + b + c)/(bc + ba + c² + ac) = 2/(a + b)

or, (ac + a² + ab + ac + bc + ab + b² + bc) = (2bc + 2ba + 2c² + 2ac)

or, (2ac + 2ab + 2bc + a² + b²) = (2bc + 2ba + 2ac + 2c²)

or, (2ac + 2ab + 2bc + a² + b² - 2bc - 2ba - 2ac) = 2c²

or, (2ac - 2ac + 2ab - 2ab + 2bc - 2bc + a² + b²) = 2c²

or, a² + b² = 2c² (Proved)

Step-by-step explanation: