Let $E$ be the midpoint of arc $BAC$. We claim that this point is the common intersection point of the two lines and the circle. Note that since $K$ lies on the angle bisector of $\angle IAJ$ and $IK=KJ$, this means that $K$ is the midpoint of arc $IJ$ in $(AIJ)$. We show that $E$ lies on the circle $(AIKJ)$. [asy][asy] size(8cm); pair A = dir(110); pair B = dir(210); pair D = dir(330); pair C = dir(270); pair I = incenter(A,B,C); pair J = incenter(A,C,D); pair K = intersectionpoints(circumcircle(A,I,J),A--C)[1]; pair P = incenter(A,B,D); pair E = dir(90); draw(unitcircle); draw(circumcircle(A,I,J)); draw(A--B--C--D--A--cycle); draw(A--C); draw(I--K--J); draw(circumcircle(A,B,I),dashed+red); draw(circumcircle(A,C,I),dashed+red); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$D$",D,dir(D)); dot("$E$",E,dir(E)); dot("$I$",I,dir(I)); dot("$J$",J,dir(J)); dot("$K$",K,dir(K)); dot("$P$",P,dir(P)); [/asy][/asy] Let $P$ be the incenter of $\triangle ABC$. We first show that $A, B, I, P$ are concyclic. Note that \[\angle APB = 90^\circ + \frac{1}{2}\angle ADB = 90^\circ + \frac{1}{2}\angle ACB = \angle AIB.\]Similarly, we obtain that $A, D, J, P$ are concyclic. Now invert about $A$. The point $P$ is sent to some point on the $A$-angle bisector (the $A$-excenter, actually). The point $E$ is sent to the intersection of the $\angle A$ external bisector with $BD$, while $I^*$ and $J^*$ are sent to the intersections of lines $B^*P^*$ and $D^*P^*$ with the angle bisectors of $\angle B^* AP^*$ and $\angle P^* AC^*$. [asy][asy] size(8cm); pair A = dir(130); pair B = dir(210); pair D = dir(330); pair C = extension(A,incenter(A,B,D),B,D); pair E = extension(B,D,A,A+dir(90)*(C-A)); pair P = extension(A,C,B,B+dir(90)*(incenter(A,B,D)-B)); pair I = extension(B,P,A,incenter(A,B,P)); pair J = extension(E,I,D,P); draw(A--B--P--D--A--cycle); draw(A--E--D); draw(E--J,dashed); draw(A--P); draw(A--I); draw(A--J); dot("$A$",A,dir(A)); dot("$B^*$",B,dir(B)); dot("$D^*$",D,dir(D)); dot("$C^*$",C,dir(C)); dot("$I^*$",I,dir(I)); dot("$J^*$",J,dir(J)); dot("$E^*$",E,dir(E)); dot("$P^*$",P,dir(P)); [/asy][/asy] Using Menelaus' Theorem on $\triangle B^* D^* P^*$ and points $E^*, I^*, J^*$, we have \[\frac{P^* I^*}{I^* B^*} \cdot \frac{B^* E^*}{E^* D^*} \cdot \frac{D^*J^*}{J^*P^*}=\frac{P^*A}{AB^*}\cdot\frac{B^*A}{AD^*}\cdot\frac{D^*A}{AP^*}=1.\]Hence, $I^*, J^*, E^*$ are collinear, which means that $E$ lies on $(AIJ)$, which is what we want to show. Now we show that $E$ is the point diametrically opposite $K$ on this circle, from which we may also see that $E$ is the midpoint of arc $IAJ$. This follows from the fact that $\angle EAK=\angle EAC=90^\circ$. Finally, we may apply our earlier lemma to cyclic quadrilateral $AIKJ$ again to see that $E$ lies on the circle $(AMN)$, and therefore the perpendicular to $IK$ at $I$ and the perpendicular to $DC$ at $D$ coincide at the $E$ which lies on the circle $(AMN)$. [asy][asy] size(10cm); pair A = dir(110); pair B = dir(210); pair D = dir(330); pair C = dir(270); pair I = incenter(A,B,C); pair J = incenter(A,C,D); pair K = intersectionpoints(circumcircle(A,I,J),A--C)[1]; pair P = incenter(A,B,D); pair E = dir(90); pair M = incenter(A,I,K); pair N = incenter(A,J,K); draw(unitcircle); draw(circumcircle(A,I,J)); draw(I--A--J); draw(A--B--C--D--A--cycle); draw(A--C); draw(I--K--J); draw(circumcircle(A,M,N),dashed+blue); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$D$",D,dir(D)); dot("$E$",E,dir(E)); dot("$I$",I,dir(I)); dot("$J$",J,dir(J)); dot("$K$",K,dir(K)); dot("$P$",P,dir(P)); dot("$M$",M,dir(180)); dot("$N$",N,dir(N)); [/asy][/asy]