If \(\frac{1}{y+z}+\frac{1}{z+x} = \frac{2}{x+y}\) then what is (x2 +

1.	If \(\frac{1}{y+z}+\frac{1}{z+x} = \frac{2}{x+y}\) then what is (x2 + y2) equal to ?
Answer» \(\frac{1}{y+z}+\frac{1}{z+x} = \frac{2}{x+y}\) ⇒ \(\frac{1}{y+z}+\frac{1}{x+z} \) = \(\frac{1}{x+y}+\frac{1}{z+x} \) ⇒ \(\frac{(x+y)-(y+z)}{(y+z)(x+y)}\) = \(\frac{(z+x)-(x+y)}{(x+y)(z+x)}\) ⇒ \(\frac{x-z}{y+z} = \frac{z-y}{z+x}\) ⇒ (x-z)(x+z) = (z-y)(z+y) ⇒ x² – z² = z² – y² ⇒ x² + y² = 2z².

If \(\frac{1}{y+z}+\frac{1}{z+x} = \frac{2}{x+y}\) then what is (x2 + y2) equal to ?

Answer»

\(\frac{1}{y+z}+\frac{1}{z+x} = \frac{2}{x+y}\)

⇒ \(\frac{1}{y+z}+\frac{1}{x+z} \) = \(\frac{1}{x+y}+\frac{1}{z+x} \) ⇒ \(\frac{(x+y)-(y+z)}{(y+z)(x+y)}\) = \(\frac{(z+x)-(x+y)}{(x+y)(z+x)}\)

⇒ \(\frac{x-z}{y+z} = \frac{z-y}{z+x}\) ⇒ (x-z)(x+z) = (z-y)(z+y)

⇒ x² – z² = z² – y² ⇒ x² + y² = 2z².