Đặt $\dfrac{a}{2019} = \dfrac{b}{2021} = \dfrac{c}{2023} = k$ Suy ra : $a = 2019k$ $b = 2021k$ $c = 2023k$ Khi đó : $\dfrac{(a-c)^2}{4} = \dfrac{(2019k - 2023k)^2}{4} = 4k^2$ $(a-b)(b-c) = (2019k - 2021k)(2021k - 2023k) = (-2k)(-2k) = 4k^2$ $\Rightarrow \dfrac{(a-c)^2}{4} = (a-b)(b-c)$