Let $a$ be the horizontal distance from $A$ to $B$ and let $b$ be the vertical distance from $A$ to $B$. By Pythagorean Theorem, $a^2+b^2=1$. Furthermore, let $\angle{BAC}=\angle{DAC}=\theta$ and $x$ be any of the three equal parts of the diagonal. Then $\cos{\theta}=\frac{1}{2x}$. But by the Law of Cosines on either $\triangle{ADC}$ or $\triangle{ABC}$, we get: $1^2=1^2+\left(3x\right)^2-2\left(1\right)\left(3x\right)\cos{\theta} \rightarrow 9x^2-6\cancel{x}\left(\frac{1}{2\xcancel{x}}\right)=0 \rightarrow 9x^2-3=0 \rightarrow x^2=\frac{1}{3}$. Let $E$ be a point outside of the rhombus such that $BE=a$ and $AE=b$. Then $\triangle{AEB}$ is right, and: $\left(3x\right)^2=\left(1+a\right)^2+b^2 \rightarrow a^2+b^2+2a+1=9x^2 \rightarrow 2a+2=9x^2 \rightarrow 2a+2=9\left(\frac{1}{3}\right) \rightarrow 2a+2=3 \rightarrow a=\frac{1}{2}$. Thus, $\left(\frac{1}{2}\right)^2+b^2=1 \rightarrow b^2=\frac{3}{4} \rightarrow b=\frac{\sqrt{3}}{2}$, since distance is clearly positive. However, $b$ is also the altitude of the rhombus, so this is our answer.