Find the segment "DC" in the obtuse triangle below

Question

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$

Answer 1

Let $\angle BAD = \angle CAD = \alpha$ then $\angle BAD =\angle BCA = 2\alpha$

Draw $DE$ such that $AE=QE$ ($E$ is the midpoint of $AQ$)

then $AE=QE=DE=5$ and $\angle ADE = \angle DAE = \alpha$

then $\angle DEC =2\alpha$ (exterior $\angle$ of $\triangle ADE$)

then $\angle DEC = \angle DCE = 2\alpha$

$DC=DE=5$ (sides are $=$ opposite $= \angle$'s)

Answer 2

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.

Answer 3

Using your result of $\measuredangle CQD = 90^{\circ} + \theta$ and the Law of sines with $\triangle CDQ$ gives

$$\begin{equation}\begin{aligned} \frac{\lvert DC\rvert}{\sin(90^{\circ} + \theta)} & = \frac{\lvert DQ\rvert}{\sin(2\theta)} \\ \frac{\lvert DC\rvert}{\cos\theta} & = \frac{\lvert DQ\rvert}{2\sin\theta\cos\theta} \\ \lvert DC\rvert & = \frac{\lvert DQ\rvert}{2\sin\theta} \end{aligned}\end{equation}$$

From $\triangle ADQ$, we get

$$\sin\theta = \frac{\lvert DQ\rvert}{\lvert AQ\rvert} \;\;\to\;\; \lvert DQ\rvert = 10\sin\theta$$

Substituting this into the earlier equation results in

$$\lvert DC\rvert = \frac{10\sin\theta}{2\sin\theta} = 5$$

However, this doesn't match your stated answer. In general, $\lvert DC\rvert$ is half that of $\lvert AQ\rvert$, as your diagram also roughly indicates.