ta có
\(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)^2\)
\(< =>a^2x^2+a^2y^2+b^2x^2+b^2y^2=a^2x^2+2axby+b^2y^2\)
<=> \(a^2y^2+b^2x^2=2axby\)
<=> \(a^2y^2+b^2x^2-2axby=0\)
<=> \(\left(ax-bx\right)^2=0\)
<=> \(ay-bx=0\)
<=> \(ay=bx\)
<=> \(\dfrac{a}{x}=\dfrac{b}{y}\)(đpcm)