Solution:

$\sqrt{2-\sqrt{2}}$ units

Refer to Figure 7. Let $\theta$ be the central angle subtended by the chord of length $\sqrt{2}$, and $\alpha$ the central angle subtended by each of the chords of equal lengths (and let $x$ be this common length). By the Law of Cosines, we have
$$x^{2}=1^{2}+1^{2}-2 \cos \alpha=2-2 \cos \alpha$$
Since $\theta+2 \alpha=180^{\circ}$, we get
$$\cos \alpha=\cos \left(90^{\circ}-\frac{\theta}{2}\right)=\sin \frac{\theta}{2}=\frac{\sqrt{2}}{2}$$
We can then solve for $x$.

Figure 7: Problem 60.1.