Có \(\dfrac{a}{b}=\dfrac{c}{d}< =>\dfrac{a}{c}=\dfrac{b}{d}\)
Áp dụng dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\)
<=> \(\left(\dfrac{a}{c}\right)^2=\left(\dfrac{b}{d}\right)^2=\left(\dfrac{a+b}{c+d}\right)^2\)
<=> \(\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\left(\dfrac{a+b}{c+d}\right)^2\)(1)
Có \(\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}\)
Áp dụng DTSBN ta có:
\(\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\dfrac{a^2+b^2}{c^2+d^2}\)(2)
Từ (1) (2) => đpcm.