Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
=>a=bk ; c=dk
=>\(\left(\dfrac{a+b}{c+d}\right)^2=\left(\dfrac{bk+b}{dk+d}\right)^2=\dfrac{b^2.\left(k+1\right)^2}{d^2.\left(k+1\right)^2}=\dfrac{b^2}{d^2}\left(1\right)\)
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}=\dfrac{\left(b^2\cdot k^2\right)+b^2}{\left(d^2\cdot k^2\right)+d^2}=\dfrac{b^2\cdot\left(k^2+1\right)}{d^2\cdot\left(k^2+1\right)}=\dfrac{b^2}{d^2}\left(2\right)\)
Từ (1) và (2) =>\(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)