Đặt \(AB=2a;BC=a\)Tọa độ \(A\left(0;0\right);B\left(2a;0\right);C\left(2a;a\right);D\left(0;a\right);N\left(x;a\right)\)\(\overrightarrow{AC}=2a;a\Rightarrow\overrightarrow{n_p}=\left(-a;2a\right)=-a\left(1;-2\right)\)\(\Rightarrow\left(AC\right):\left(x-0\right)-2\left(y-0\right)=0\)\(\Rightarrow\left(AC\right):y=\dfrac{1}{2}x\)Hệ số góc của phương trình đường thẳng \(BN\) là \(\dfrac{a-0}{x-2a}=\dfrac{a}{x-2a}\)Ta có \(AC\perp BN\)\(\Leftrightarrow\dfrac{a}{x-2a}.\dfrac{1}{2}=-1\)\(\Leftrightarrow\dfrac{a}{2\left(x-2a\right)}=-1\)\(\Leftrightarrow a=-2\left(x-2a\right)\)\(\Leftrightarrow x=\dfrac{3a}{2}\)\(\Rightarrow N\left(\dfrac{3a}{2};a\right)\)\(DN=\sqrt{\left(\dfrac{3a}{2}-0\right)^2-\left(a-a\right)^2}=\dfrac{3a}{2}\)\(CN=\sqrt{\left(\dfrac{3a}{2}-2a\right)^2+\left(a-a\right)^2}=\dfrac{a}{2}\)\(\Rightarrow\dfrac{DN}{CN}=\dfrac{\dfrac{3a}{2}}{\dfrac{a}{2}}=3\)Câu trả lời: Đặt \(AB=2a;BC=a\)Tọa độ \(A\left(0;0\right);B\left(2a;0\right);C\left(2a;a\right);D\left(0;a\right);N\left(x;a\right)\)\(\overrightarrow{AC}=2a;a\Rightarrow\overrightarrow{n_p}=\left(-a;2a\right)=-a\left(1;-2\right)\)\(\Rightarrow\left(AC\right):\left(x-0\right)-2\left(y-0\right)=0\)\(\Rightarrow\left(AC\right):y=\dfrac{1}{2}x\)Hệ số góc của phương trình đường thẳng \(BN\) là \(\dfrac{a-0}{x-2a}=\dfrac{a}{x-2a}\)Ta có \(AC\perp BN\)\(\Leftrightarrow\dfrac{a}{x-2a}.\dfrac{1}{2}=-1\)\(\Leftrightarrow\dfrac{a}{2\left(x-2a\right)}=-1\)\(\Leftrightarrow a=-2\left(x-2a\right)\)\(\Leftrightarrow x=\dfrac{3a}{2}\)\(\Rightarrow N\left(\dfrac{3a}{2};a\right)\)\(DN=\sqrt{\left(\dfrac{3a}{2}-0\right)^2-\left(a-a\right)^2}=\dfrac{3a}{2}\)\(CN=\sqrt{\left(\dfrac{3a}{2}-2a\right)^2+\left(a-a\right)^2}=\dfrac{a}{2}\)\(\Rightarrow\dfrac{DN}{CN}=\dfrac{\dfrac{3a}{2}}{\dfrac{a}{2}}=3\)