Nếu đặt AM là x thì $DM=\dfrac{x}{3}$, $AD=\dfrac{2x}{3}$
Và $BD=\sqrt{x^2-(\dfrac{x}{3})^2}=\dfrac{\sqrt{8}x}{3}$
$DN=\dfrac{\sqrt{8}x}{3}:2=\dfrac{\sqrt[]{2}x}{3}$
Tam giác ADN vuông tại D áp dụng py-ta-go:
$AN=\sqrt[]{(\dfrac{2x}{3})^2+(\dfrac{\sqrt[]{2}x}{3})^2}=\dfrac{\sqrt[]{6}x}{3}$
\Rightarrow $AC=\dfrac{\sqrt[]{24}x}{3}$
$AB=\sqrt[]{(\dfrac{\sqrt{8}x}{3})^2+(\dfrac{2x}{3})^2}= \dfrac{\sqrt[]{12}x}{3}$
$AC:AB= \dfrac{\dfrac{\sqrt[]{24}x}{3}}{\dfrac{\sqrt[]{12}x}{3}}=\sqrt[]{2}$