Find the segment "DC" in the obtuse triangle below

In an obtuse triangle $ABC$, obtuse at $B$, the internal bisector $AD$ is drawn and in $AC$ the point $"q"$ is taken such that $m∡𝐴⁢𝐷⁢𝑄=90^𝑜$.Calculate $DC$. If: $AQ = 10$ and $AB = BC$.$Answer:10$

I try

$\angle QAD= \theta = \angle DAB$

$\frac{AB}{BD} = \frac{10+CQ}{CD}$

$\triangle ABC_{(isosc.)}\implies \angle C = 2\theta$

$\therefore \angle A = 180^o -4\theta$

$ \angle BDA = 3\theta \implies \angle QDC = 90-3\theta$

$ \angle CDQ = 90+\theta$