Let $R$ be the midpoint of $AQ$, so $\lvert AR\rvert = \lvert RQ\rvert = 5$. Since $\triangle ADQ$ is right-angled, then $AQ$ is the diameter of its circumcircle, so $R$ is its center, which means that $\lvert DR\rvert = \frac{10}{2} = 5$.
Since $\triangle ARD$ is isosceles, then $\measuredangle ADR = \theta$, so $\measuredangle RDQ = \theta$ also. Using that $\measuredangle BDA = 3\theta$, we get
$$\measuredangle RDC = 180^{\circ} - 3\theta - \theta = 180^{\circ} - 4\theta$$
With $\measuredangle DCR = 2\theta$, this means
$$\measuredangle DRC = 180^{\circ} - (180^{\circ} - 4\theta) - 2\theta = 2\theta$$
This shows $\triangle RDC$ is also isoscles, so
$$\lvert DC \rvert = \lvert DR \rvert = 5$$
which matches the result, where the law of sines was used instead, of my other answer.