Let $P, Q$ be the circumcenters of the triangles $\triangle ABD, \triangle AMN$ and $(O)$ a circle with diameter $AC$ centered at the midpoint $O$ of this diagonal of the quadrilateral $ABCD$. Let the lines $CB , CD, CM, CN$ intersect the circle $(O)$ at points $B', D', M', N'$ different from the point $C$. The angles $\angle AB'B \equiv \angle AB'C$, $\angle AD'D \equiv \angle AD'C$, $\angle AM'M \equiv \angle AM'C$, $\angle AN'N \equiv \angle AN'C$ are all right, because they span the diameter $AC$ of the circle $(O)$. The angles $\angle ABB', \angle ADD', \angle AMM', \angle ANN'$ are all equal to $45^o$: $\angle ABB' = 180^o - \angle ABD = 180^o - 135^o = 45^o$ $\angle ADD' = 180^o - \angle ADC = 180^o - 135^o = 45^o$ $\angle AMM' = \angle ADC - \angle MCD = 135^o - 90^o = 45^o$ $\angle ANN' = \angle ABC - \angle NCB = 135^o - 90^o = 45^o$ Hence, the right angle triangles $\triangle ABB', \triangle ADD', \triangle AMM', \triangle ANN'$ are all isosceles and consequently, the perpendicular bisectors of the segments $AB, AD, AM, AN$ pass through the points $B', D', M', N'$. Let the perpendicular bisectors of the segments $AB, AD, AM, AN$ meet the circle $(O)$ at points $E, F, E', F'$. The angles $\angle AB'E = \angle AM'E' = 45^o$ are equal, hence, they span the same arc $AE \equiv AE'$ of the circle $(O)$ and the points $E \equiv E'$ are identical. Similarly, the angles $\angle AD'F = \angle AN'F' = 45^o$ are equal, hence, they span the same arc $AF \equiv AF'$ of the circle $(O)$ and the points $F \equiv F'$ are identical. Since these angles are all equal to $45^o$, the points $E, F$ are diametrally opposite points of the circle $(O)$, the points $E, O, F$ are collinear and $OE = OF$. The perpendicular bisectors $B'E \parallel N'F$ of the segments $AB, AN$ are parallel, being both perpendicular to the same line $AB \equiv AN$ and they pass through the circumcenters $P, Q$, respectively. Similarly, the perpendicular bisectors $D'F \parallel M'F$ of the segments $AD, AM$ are parallel, being both perpendicular to the same line $AD \equiv AM$ and they also pass through the circumcenters $P, Q$, respectively. The quadrilateral $PEQF$ is a parallelogram and the center $O$ of the circle $(O)$ is the intersection of its diagonals, being the midpoint of the diagonal $EF$. Consequently, the centers of the circles $(P), (O), (Q)$ are collinear. Since these 3 circles intersect at the point $A$, they belong to the same elliptic pencil and they all intersect at one other point, the intersection $K$ of the circles $(P), (Q)$. Since the angles $\angle AKC$ spans the diameter of the circle $(O)$, it is right and $AK \perp KC$.

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http://www.mathlinks.ro/Forum/viewtopic.php?t=16899 [Moderator edit: No, this is not the same problem as the one discussed on http://www.mathlinks.ro/Forum/viewtopic.php?t=16899 . In fact, in this problem, the points M and N lie on the extensions of the lines AD and AB and satisfy < MCD = < NCB = 90°, while in the problem from http://www.mathlinks.ro/Forum/viewtopic.php?t=16899 , the points M and N lie on the extensions of the lines AB and AD and satisfy < MCD = < NCB = 90°. Funnily enough, the two problems are different but both correct.]

Let $P,Q$ be the midpoints of $BM,DN$, from $KBN \sim KDM$ we get $KBQ \sim KDP$, then $A,K,P,Q(,C)$ concyclic and we are done.

the quadraterial ABCD is a parallelogram. So, <BAC = <BCD = 45, <DCN = <ADC - <DCN = 135 - 90 = 45 similarly, <MCB = 45 <MCN = <MCB + <BCN = 45+90 = 135 <MCN + <MAN = 135 + 45 = 180, so, A,K,N,M,C are concyclic. <AMC = 180 - <MCD = 180 - 90 = 90, <AKC = 180 - <AMC = 180 - 90 = 90

I added another solution to here of this beautiful problem.

Let F and E be midpoints of BN and DM. ∠KBN = ∠KDM and ∠KMD = ∠KNB ---> KNB and KMD are similar ---> ∠KFA = ∠KEA ---> AKFE is cyclic. ∠CBN = 45 and ∠BCN = 90 ---> ∠CFA = 90 ∠CDM = 45 and ∠DCM = 90 ---> ∠CEA = 90 so AFCE is cyclic as well and we have ∠AKC = ∠AFC = 90 we're Done.

liekkas wrote: Let $P,Q$ be the midpoints of $BM,DN$, from $KBN \sim KDM$ we get $KBQ \sim KDP$, then $A,K,P,Q(,C)$ concyclic and we are done. Hello, shouldn't it be /{BN} and DQ?