Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Do đó ta có:
\(\dfrac{ac}{bd}=\dfrac{bkdk}{bd}=k^2\left(1\right)\)
Ta cũng có:
\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}\)
\(=\dfrac{b^2k^2+d^2k^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\left(2\right)\)
Từ (1) và (2)
\(\Rightarrow\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}=\left(k^2\right)\)\(\Rightarrowđpcm\)
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