Ta có: \(\frac{a}{b}=\frac{c}{d}\)=>\(\frac{a}{c}=\frac{b}{d}\)
=>\(\left(\frac{a}{c}\right)^2=\left(\frac{b}{d}\right)^2=\frac{ab}{cd}\)
=>\(\frac{ab}{cd}=\frac{a^2}{c^2}=\frac{b^2}{d^2}\)
=>\(\frac{ab}{cd}=\frac{a^2+b^2}{c^2+d^2}=\frac{a^2-b^2}{c^2-d^2}\) (theo tính chất dãy tỉ số bằng nhau)
=>\(\frac{a^2+b^2}{a^2-b^2}=\frac{c^2+d^2}{c^2-d^2}\)(đpcm)