mathisreal wrote: Let $ABCD$ be a convex quadrilateral, where $R$ and $S$ are points in $DC$ and $AB$, respectively, such that $AD=RC$ and $BC=SA$. Let $P$, $Q$ and $M$ be the midpoints of $RD$, $BS$ and $CA$, respectively. If $\angle MPC + \angle MQA = 90$, prove that $ABCD$ is cyclic. Let the reflection of $C$ across point $P$ be $K_1$. It's easy to observe that $MP// AK_1$. Moreover, $AD=RC=DK_1$. Thus, $\angle ADC=2\cdot \angle AK_1C=2\cdot \angle MPC$ Likewisely, we will obtain that $\angle ABC=2\cdot \angle MQA$. Hence, $\angle ADC+\angle ABC=2\cdot (\angle MPC+\angle MQA)=180^\circ $ In other words, $ABCD$ is concyclic. Done

Nice problem. Let $X$ and $Y$ be the midpoints of $DC$ and $AB$ respectively. From the conditions (i), (ii) we get that $PX = \frac{AD}{2} = XM$ and similarly $QY = YM$. Now from condition (iii) and as $MX||AD$, $MY||BC$, \[\angle ADC + \angle CBA =\angle MXC + \angle MYA = 2\cdot MPC + 2\cdot \angle MQA = 2\cdot 90^{\circ} = 180^{\circ}.\]Proved. [asy][asy] import graph; size(5cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -11.82, xmax = 12.58, ymin = -3.34, ymax = 9.78; /* image dimensions */ pair A = (-5.,0.), B = (6.,0.), C = (5.88,6.76), D = (-4.,8.), R = (-2.1195007509218162,7.763985924204762), Y = (0.5,0.), X = (0.94,7.38), M = (0.44,3.38), P = (-3.059750375460908,7.881992962102381), Q = (3.880532502432124,0.); /* draw figures */ draw(A--B, linewidth(1.)); draw(B--C, linewidth(1.)); draw(C--D, linewidth(1.)); draw(D--A, linewidth(1.)); draw(A--C, linewidth(1.)); draw(Y--M, linewidth(1.)); draw(M--X, linewidth(1.)); draw(D--R, linewidth(1.)); draw((1.7610650048642482,0.)--B, linewidth(1.)); draw(M--Q, linewidth(1.)); draw(M--P, linewidth(1.)); /* dots and labels */ dot(A,dotstyle); label("$A$", A, dir(-135)); dot(B,dotstyle); label("$B$", B, dir(0)); dot(C,dotstyle); label("$C$", C, dir(0)); dot(D,dotstyle); label("$D$", D,dir(150)); dot(R,linewidth(2.pt) + dotstyle); label("$R$", R,dir(90)); dot((1.7610650048642482,0.),linewidth(2.pt) + dotstyle); label("$S$", (1.68,-0.8), NE * labelscalefactor); dot(Y,linewidth(4.pt) + dotstyle); label("$Y$",Y,dir(-90)); dot(X,linewidth(4.pt) + dotstyle); label("$X$", X,dir(90)); dot(M,linewidth(4.pt) + dotstyle); label("$M$", M,dir(180)); dot(P,linewidth(2.pt) + dotstyle); label("$P$", P, dir(120)); dot(Q,linewidth(2.pt) + dotstyle); label("$Q$", Q,dir(-90)); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

Really nice problem! The conditions given make you draw an extra segment to complete the proof, indeed, beautiful problem! If you get a point C' such that C'D=CR, you finish the problem.