\(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\) <=> \(\frac{a^2+b^2}{c^2+d^2}=\frac{2ab}{2cd}=\frac{a^2+b^2+2ab}{c^2+d^2+2cd}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\) (1)
\(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\) <=> \(\frac{a^2+b^2}{c^2+d^2}=\frac{2ab}{2cd}=\frac{a^2+b^2-2ab}{c^2+d^2-2cd}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\) (2)
Từ (1) và (2) suy ra: \(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\left(\frac{a-b}{c-d}\right)^2\)
=> \(\frac{a+b}{c+d}=\frac{a-b}{c-d}\)
Ta có: \(\frac{a+b}{c+d}=\frac{a-b}{c-d}=\frac{a+b+a-b}{c+d+c-d}=\frac{2a}{2c}=\frac{a}{c}\)
\(\frac{a+b}{c+d}=\frac{a-b}{c-d}=\frac{a+b-a+b}{c+d-c+d}=\frac{2b}{2d}=\frac{b}{d}\)
=> \(\frac{a}{c}=\frac{b}{d}\) hay \(\frac{a}{b}=\frac{c}{d}\)
