The proof of $\dfrac{a^2+b^2}{a^2-b^2} = \dfrac{ac+bd}{ac-bd}$ given $\dfrac{a}{b} = \dfrac{c}{d}$ uses the substitution

Question

The proof of $\dfrac{a^2+b^2}{a^2-b^2} = \dfrac{ac+bd}{ac-bd}$ given $\dfrac{a}{b} = \dfrac{c}{d}$ uses the substitution—

Accepted Answer

$a = bk,\ c = dk$

Answer

$a + b = c + d$

Answer

$abcd = 1$

Answer

$a = c$

The proof of a2+b2a2−b2=ac+bdac−bd\dfrac{a^2+b^2}{a^2-b^2} = \dfrac{ac+bd}{ac-bd}a2−b2a2+b2​=ac−bdac+bd​ given ab=cd\dfrac{a}{b} = \dfrac{c}{d}ba​=dc​ uses the substitution—