In triangle $$ABC$$, the incircle, with center $$I$$, touches the side $$BC$$ at point $$D$$. Line $$DI$$ meets $$AC$$ at $$P$$. The tangent line from $$P$$ to the incircle (different from $$AC$$) intersects $$AB$$ at $$Q$$. If line $$QI$$ and $$BC$$ intersect at point $$R$$, prove that quadrilateral $$AIRC$$ is a cyclic.

Let $$\angle BAC = \alpha$$, $$\angle ABC = \beta$$ and $$\angle ACB = \gamma$$. Since the incircle $$\omega$$ touches $$BC$$ at $$D$$, we have $$PD \perp BC$$, so in right triangle $$PDC$$, $$\angle DPC = 90^\circ - \gamma$$. As $$I$$ lies on $$PD$$, line $$PD$$ bisects the angle between the two tangents $$PC$$ and $$PQ$$ drawn from $$P$$ to $$\omega$$, so $$\angle QPD = \angle DPC = 90^\circ - \gamma$$, giving $$\angle QPC = 180^\circ - 2\gamma$$ and $$\angle APQ = 2\gamma$$. In $$\triangle APQ$$ the exterior angle at $$Q$$ is $$\angle BQP = \alpha + 2\gamma$$, and since $$\omega$$ is the $$A$$-excircle of $$\triangle APQ$$, ray $$QI$$ bisects it, so $$\angle BQR = \frac{\alpha}{2} + \gamma$$. Then in $$\triangle BQR$$, $$\angle BRQ = 180^\circ - \beta - \frac{\alpha}{2} - \gamma = \frac{\alpha}{2}$$. Since $$AI$$ bisects $$\angle BAC$$, $$\angle IAC = \frac{\alpha}{2}$$, so $$\angle IAC + \angle IRC = \frac{\alpha}{2} + \left(180^\circ - \frac{\alpha}{2}\right) = 180^\circ$$, and therefore quadrilateral $$AIRC$$ is cyclic.