Ta có \(\frac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}=\frac{a^2+c^2+2ab-2ac-2bc+\left(a-c\right)^2}{b^2+c^2+2ab-2ac-2bc+\left(b-c\right)^2}\)
\(=\frac{\left(a-c\right)^2+2b\left(a-c\right)+\left(a-c\right)^2}{\left(b-c\right)^2+2a\left(b-c\right)+\left(b-c\right)^2}=\frac{\left(a-c\right)\left(a-c+2b+a-c\right)}{\left(b-c\right)\left(b-c+2a+b-c\right)}=\frac{\left(a-c\right)\left(2a+2b-2c\right)}{\left(b-c\right)\left(2a+2b-2c\right)}=\frac{a-c}{b-c}\)
⇒điều phải chứng minh