Đặt \(\frac{a}{b}=\frac{d}{c}=k\)
=> a=bk; b=ck
Suy ra:
\(\frac{a^2+d^2}{b^2+c^2}=\frac{\left(bk\right)^2+\left(ck\right)^2}{b^2+c^2}=\frac{k^2\left(b^2+c^2\right)}{b^2+c^2}=k^2\)
\(\frac{a.d}{b.c}=\frac{bkck}{b.c}=\frac{k^2.b.c}{b.c}=k^2\)
=>\(\frac{a^2+d^2}{b^2+c^2}=\frac{ad}{bc}\)