1984 AMC 12 第 27 题

Let $AC=s$ and use the altitude-to-hypotenuse similarity relation

Then compare $\cos\angle ACB$ in right triangle $ABC$ and isosceles triangle $BDC$

Let $AC=s.$ The right-triangle projection relation gives $$AC^2=FC\cdot BC,$$ so $BC=s^2.$ Since $BD=DC=1,$ the numerator and denominator in the Law of Cosines satisfy $$\begin{aligned} BC^2+DC^2-BD^2&=s^4,\\ 2(BC)(DC)&=2s^2. \end{aligned}$$ Hence $\cos\angle BCD=\frac{s^2}{2}.$ But $D$ lies on $AC,$ and in right triangle $ABC,$ $\cos\angle BCA=\frac{AC}{BC}=\frac{1}{s}.$ Thus $\frac{s^2}{2}=\frac{1}{s},$ so $s^3=2$ and $AC=\sqrt[3]{2}.$

Therefore, the correct answer is C.