# Rhombus, circumcircle and incenter

I try to do this problem:

It is given a rhombus $$ABCD$$ with sidelength $$a$$. On the line $$AC$$ are chosen the points $$M$$ and $$N$$ in such a way that $$C$$ lies between $$A$$ and $$N$$ and $$MA\cdot NC= a^{2}$$. We denote with $$P$$ the intersection point of $$MD,BC$$ and $$Q$$ is the intersection point of $$ND,AB$$. Prove that $$D$$ is the incenter of the triangle $$PQB$$.

I do this figure

By the sinus theorem applied to the triangles $$MAD$$ and $$NCD$$ implies $$\alpha+\beta=\theta$$ and these triangles are similar.

Then I've drawn the circumcircle to the triangle $$BPQ$$.

I wonder if the points $$M$$ and $$N$$ are on this circumcircle, if they are, the inscribed angles $$NMP$$ and $$NQP$$ are equal and the prove is done.

I appreciate any help, thanks.

Since $$\angle AMD=\angle AQD$$, quadrilateral $$AMQD$$ is cyclic. Therefore $$\angle MQA=\beta$$ and $$\angle MQD=\alpha+\beta$$.
Similarly, we can prove that $$\angle NPD=\alpha+\beta$$.
Therefore $$\angle MQN=\angle MPN\implies MQPN\text{ is cyclic.}$$