This problem has already been discussed here: http://www.mathlinks.ro/Forum/viewtopic.php?t=113117

After showing $ \angle PMD =\angle QMD$ the problem is equivalent, by duality, to the following problem: Let $ PQRS$ be a cyclic quadrilateral and let: $ M'\in RS\cap QP$ $ T\in SP\cap RQ$ $ l\bot OM'$ Y $ l$ and $ M'\in l$ $ X\in SP\cap l$ $ Y\in RQ\cap l$ Show $ XM' = YM'$. (Showing $ XM' = YM'$ is equivalent to $ \angle XOM' =\angle YOM$) if $ M$ is the pole of $ l$, $ TM$,$ TX$,$ TM'$ and $ TY$ cut lines in harmonic conjugate (easely proven by Ceva and Menelao on the $ TSQ$) and $ TM\parallel l$ ($ TM\bot MM'$) so $ XM' = YM'$.

by Googo's theorem,$\angle AMD=\angle CMD$ since $DPOMQ$cyclic,PD=DQ so $\angle PMD=\angle QMD$ hence $\angle AMP=\angle CMQ$.

about Googo's theorem,see here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=425471

Nice one! Let $\varphi$ touch sides $CB$ and $AB$ at $R$ and $S$, respectively. Let $I$ be the center of $\varphi$. Obviously, $IM\perp BD$. Then $D\widehat{M}I=D\widehat{P}I=D\widehat{Q}I=\frac{\pi}{2}$, therefore $D$, $P$, $M$, $I$, $Q$ are concyclic. This yields $P\widehat{M}D=D\widehat{M}Q$, since $DP=DQ$, and now we must show that $A\widehat{M}D=C\widehat{M}D$. Invert around $\varphi$; for each point $X$ denote its inverse by $X^\ast$. Then $A^\ast$ and $C^\ast$ are the midpoints of $PS$ and $QR$, respectively. Observe that $BD$ is the polar of $K=PQ\cap RS$, therefore $M$ is the inverse of $K$. Since $IM\perp BD$, what we want to prove is equivalent to $\angle(AM, MI)=\angle(MI, MC)$, or, equivalently, $\angle(A^\ast I, A^\ast K)=\angle(C^\ast K, CI)$. Notice that $\angle(A^\ast I, PS)=\angle(QR, C^\ast I)=\frac{\pi}{2}$, therefore subtracting $\frac{\pi}{2}$ from both sides reduces the problem to showing that $\angle(PS, A^\ast K)=\angle(C^\ast K, QR)$. But this is easy: observe that triangles $KPS$ and $KRQ$ are similar and $A^\ast$ and $C^\ast$ are the midpoints of $PS$ and $QR$, respectively, and this yields the desired conclusion.

What a nice problem! Let $S,R$ be the tangency points of the incircle with $AB,BC$ respectively. Let $AC \cap BD = L$ and $I$ the center of the incircle. Furthermore let $M' = SR \cap PQ$ and $SP \cap QR = T$. Note that $BD$ is the polar of $M'$ because of the tangents and since $AC$ is the polar of $T$ by Brokard's theorem we must have $M'$ is on line $AC$. Since $IM \perp BD$ we must also have $I,M,M'$ are collinear. Brianchon's theorem on degenerate hexagon $APDQCB$ gives $AQ,PC,DB$ concurrent. Therefore $(A,C;L,M') = -1$. Thus the pencil $MA,MC,ML,MI$ is harmonic and $\angle IML = 90^{\circ}$ so $MD$ bisects $\angle AMC$. To conclude note that $\angle IMD = \angle IPD = \angle IQD = 90^{\circ}$ so $PMIQD$ is cyclic and $DP=DQ$ meaning $\angle PMD = \angle DMQ.$ Subtracting this from $\angle AMD = \angle DMC$ we get the desired $\angle AMP = \angle CMQ.$