A sketch: In my opinion, this is a pretty hard but natural problem. Just use the angle bisectors to obtain two harmonic bundles, which easily gets to conclude that $KL$, $MN$ and $AC$ concur at a point (possibly at infinity). Define $K'L'M'N'$ to be the quadrilateral formed by the tangency points of the incircle and sides $AB, BC, CD, DA$ respectively. A suitable application of Brianchon's and Pascal's theorems, and some harmonic bundles lead us to infer that $ KL, K'L', AC, M'N', MN$ concur at a point $Q$; $LM, L'M', BD, KN, K'N'$ concur at $R$ and $KM, LN, K'M', L'N', AC, BD$ concur at $P$. Using harmonic bundles again, we can show that $BD, AC$ are the $Q, R$-polars with respect to the incircle and $(KLMN)$, and from here it's not hard to argue that these circles must be concentric and finally, congruent, so $K=K'$ and the other equalities hold simultaneously. The final step is just to prove that $ABCD$ is cyclic, which is just a simple matter of angle chasing. The no-perpendicularity constraint of $AC$ and $BD$ prevents us from having the extreme case of $KL\parallel MN, \ LM\parallel LN$. The official solution includes hostile reasoning of how such a condition avoids the aforesaid parallelisms, which turns out to be trivial using harmonic bundles.

Jafet98 wrote: I've just realized that the no-perpendicularity constraint of $AC$ and $BD$ prevents us of having the extreme case of $KL\parallel MN, \ LM\parallel LN$. Well, I didn't pay attention to this issue since, when using harmonic bundles (and the real projective plane), you just add a point at infinity to each class of parallel lines and the line at infinity as well. However, the official solution includes a hostile reasoning of how such a condition avoids the aforesaid parallelisms, which turns out to be trivial using harmonic bundles. Sure, but the conclusion of the circumcircles of $KLMN$ and $K'L'M'N'$ being concentric doesn't hold when the concurrency points are ideal. In fact, the condition given in the problem trivially holds whenever $ABCD$ is a (not necessarily cyclic) kite. I do agree that the argument given in the official solution is unnecessarily complex though.

juckter wrote: Jafet98 wrote: I've just realized that the no-perpendicularity constraint of $AC$ and $BD$ prevents us of having the extreme case of $KL\parallel MN, \ LM\parallel LN$. Well, I didn't pay attention to this issue since, when using harmonic bundles (and the real projective plane), you just add a point at infinity to each class of parallel lines and the line at infinity as well. However, the official solution includes a hostile reasoning of how such a condition avoids the aforesaid parallelisms, which turns out to be trivial using harmonic bundles. Sure, but the conclusion of the circumcircles of $KLMN$ and $K'L'M'N'$ being concentric doesn't hold when the concurrency points are ideal. In fact, the condition given in the problem trivially holds whenever $ABCD$ is a (not necessarily cyclic) kite. I do agree that the argument given in the official solution is unnecessarily complex though. I agree with you. I'm now left to wait the final results.

Let \(E=\overline{AB}\cap\overline{CD}\), \(X=\overline{AB}\cap\overline{CD}\), \(Y=\overline{AD}\cap\overline{BC}\). Also let the incircle of \(ABCD\) touch \(\overline{AB}\), \(\overline{BC}\), \(\overline{CD}\), \(\overline{DA}\) at \(K'\), \(L'\), \(M'\), \(N'\). I claim \(K=K'\), etc. Claim: \(\overline{KL}\cap\overline{MN}\) is the harmonic conjugate of \(E\) wrt.\ \(\overline{AC}\). (In particular, it lies on \(\overline{AC}\).) Proof. Let \(K^*=\overline{AB}\cap\overline{LN}\) and \(L^*=\overline{BC}\cap\overline{KM}\). From the angle bisectors we have \(-1=(AB;KK')=(CB;LL')\), so \(\overline{AC}\), \(\overline{KL}\), \(\overline{K^*L^*}\) concur at a point \(F\). But \[-1=(AB;KK^*)\stackrel{L^*}=(AC;EF).\]The claim readily follows by symmetry. \(\blacksquare\) [asy][asy] size(8cm); defaultpen(fontsize(10pt)); pen pri=red; pen sec=orange; pen tri=fuchsia; pen fil=invisible; pen sfil=invisible; pen tfil=invisible; pair I, K, L, M, NN, A, B, C, D, EE, F, G, X, Y, Kp, Lp; I=(0, 0); K=dir(120); L=dir(40); M=dir(270); NN=2*foot(I, L, foot(L, K, M))-L; A=2*(NN+K)/length(NN+K)^2; B=2*(K+L)/length(K+L)^2; C=2*(L+M)/length(L+M)^2; D=2*(M+NN)/length(M+NN)^2; EE=extension(K, M, L, NN); F=extension(K, L, M, NN); G=extension(K, NN, L, M); X=extension(A, B, C, D); Y=extension(A, D, B, C); Kp=extension(A, B, L, NN); Lp=extension(B, C, K, M); filldraw(circumcircle(A, B, C), fil, pri); fill(A -- B -- C -- D -- cycle, fil); filldraw(circle(I, 1), sfil, sec); fill(K -- L -- M -- NN -- cycle, sfil); draw(F -- L -- M -- F, sec); draw(L -- M -- NN -- K, sec); draw(X -- B -- C -- X, pri); draw(Lp -- C -- D -- Y, pri); draw(Y -- X, tri); draw(F -- C, tri); draw(B -- D, tri); draw(Lp -- M, sec); draw(Kp -- L, sec); draw(Kp -- Lp,sec); dot("\(A\)", A, dir(100)); dot("\(B\)", B, dir(70)); dot("\(C\)", C, SE); dot("\(D\)", D, SW); dot("\(K\)", K, dir(75)); dot("\(L\)", L, NE); dot("\(M\)", M, S); dot("\(N\)", NN, dir(240)); dot("\(X\)", X, SW); dot("\(Y\)", Y, NE); dot("\(F\)", F, NW); dot("\(E\)", EE, dir(295)); dot("\(K^*\)", Kp, dir(300)); dot("\(L^*\)", Lp, N); [/asy][/asy] Claim: \(E=\overline{K'M'}\cap\overline{L'N'}\). Proof. Brianchon theorem on \(AK'BCM'D\), \(ABL'CDN'\). \(\blacksquare\) Claim: \(\overline{K'L'}\cap\overline{M'N'}\) is the harmonic conjugate of \(E\) wrt.\ \(\overline{AC}\). (In particular, it lies on \(\overline{AC}\).) Proof. By Pascal theorem on \(K'K'L'N'N'M'\), the point \(F=\overline{K'L'}\cap\overline{M'N'}\) lies on \(\overline{AEC}\). Also let \(G=\overline{K'N'}\cap\overline{L'M'}\). Note that \[-1=G(K'L';F'E')=(AC;EF),\]where the second equality is by taking the dual wrt.\ \((K'L'M'N')\). \(\blacksquare\) Remark: By Pascal theorem on \(K'K'L'M'M'N'\) and \(K'L'L'M'N'N'\), points \(X\), \(Y\) lie on \(\overline{FG}\). Claim: \(K=K'\), \(L=L'\), \(M=M'\), \(N=N'\). Proof. We have proven via Claims 2 and 3 that \((KLMN)\) and \((K'L'M'N')\) coincide as the polar circle of \(\triangle EXY\). But \((K'L'M'N')\) is tangent to each side of \(ABCD\), so \(K=K'\) is unique, etc. \(\blacksquare\) Finally, \(\overline{KM}\cap\overline{LN}\) readily implies \(ABCD\) is bicentric; to spell it out, \[\angle BAD=180^\circ-\widehat{NK}=\widehat{LM}=180^\circ-\angle DCB,\]as needed.