Đặt:
\(\dfrac{a}{b}=\dfrac{c}{d}=t\)\(\Leftrightarrow\left\{{}\begin{matrix}a=bt\\c=dt\end{matrix}\right.\)
Khi đó:
\(\left(\dfrac{a+b}{c+d}\right)^2=\left(\dfrac{bt+b}{dt+d}\right)^2==\left[\dfrac{b\left(t+1\right)}{d\left(t+1\right)}\right]^2=\dfrac{b^2}{d^2}\)
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{b^2t^2+b^2}{d^2t^2+d^2}=\dfrac{b^2\left(t^2+1\right)}{d^2\left(t^2+1\right)}=\dfrac{b^2}{d^2}\)
Ta có điều phải chứng minh
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\)
nên \(\left(\dfrac{a}{c}\right)^2=\left(\dfrac{b}{d}\right)^2=\left(\dfrac{a+b}{c+d}\right)^2\)
suy ra \(\dfrac{a^2+b^2}{c^2+d^2}=\left(\dfrac{a+b}{c+d}\right)^2\Rightarrow dpcm\)
