Answer by Lion Heart for Find the segment "DC" in the obtuse triangle below

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)