Let $\angle NCB=\alpha\ ; \angle CNP=\beta$ $\angle BCQ=\frac{1}{2}\angle AQP \leftrightarrow 180^0-\angle BQP=2(90^o-\angle BQC) \leftrightarrow \angle NQP=2\angle NQC$ To $\triangle NQP$ and point $C$ we use Ceva's:$\frac{sin \angle CQP}{sin \angle CQN} \cdot \frac{cos \alpha}{sin \beta} \cdot \frac{cos \frac{\beta}{2}}{cos(\alpha -\frac{\beta}{2})}=1$ and $\frac{cos \alpha}{sin \beta} \cdot \frac{cos \frac{\beta}{2}}{cos\alpha -\frac{\beta}{2}}=\frac{cos\alpha}{2sin\frac{\beta}{2}cos(\alpha -\frac{\beta}{2})}=\frac{cos\alpha}{sin(\beta -\alpha)+sin\alpha}=\frac{1}{tan\alpha+\frac{sin(\beta -\alpha)}{cos\alpha}}$ Let point $H$we choose satisfy $HN \perp AB ; HP \perp AD$ and let the length of $AB$=1 So $tan\alpha+\frac{sin(\beta -\alpha)}{cos\alpha}=NB+\frac{\frac{PH}{NC}}{\frac{1}{NC}}=1 \rightarrow \frac{sin \angle CQP}{sin \angle CQN}=1 \rightarrow \angle CQP=\angle NQP \rightarrow \angle NQP=2\angle NQC$ $Q.E.D$

I haven't found a graceful proof

Just notice $PQ$ is tangent to the circle $(C,CB)$. Best regards, sunken rock

Dear Mathlinkers, http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=584122 http://jl.ayme.pagesperso-orange.fr/Docs/Miniatures%20Geometriques%20addendum%20II.pdf p. 18-20. Sincerely Jean-Louis