$I$ and $J$ are the incenter and $A$-excenter in $ABC$. $K$ and $L$ are similarly defined in $ACD$. $IL \cap JK$ lies on the bisector of $\angle BCD$. Problem 96 (India Postals 2015). Let $ABCD$ be a convex quadrilateral. In $\triangle ABC$, let $I$ and $J$ be the incenter and $A$-excenter, respectively. In $\triangle ACD$, let $K$ and $L$ be the incenter and $A$-excenter, respectively. Show that the lines $IL$, $JK$, and the bisector of $\angle BCD$ are concurrent.

I came across this problem in A Beautiful Journey Through Olympiad Geometry by Stefan Lozanovski. My thoughts to solving this problem ranged from using Menelaus's/Ceva's theorem (trigonometric form, perhaps?) to using Desargues's theorem to convert the concurrency into a collinearity, though I don't immediately see how to make meaningful progress in either approaches. After being stuck for a while, I decided to look it up and found this thread, which explicitly asked for an analytic solution. To quote the author himself,
"The word 'Beautiful' in the book’s title means that we will explore only synthetic approaches and proofs, which I find elegant and beautiful. We will not see any analytic approaches, such as Cartesian or barycentric coordinates, nor we will do complex number or trigonometry bashing."
That's why I expect there to indeed be a geometrical solution. For those who do not mind the clutter, the image below is an attempt using Desargues's Theorem, where $R$ is a point chosen on the bisector of $\angle BCD$ such that $P ∈ AD$. The triangles I assumed perspectivity are $\triangle ICJ$ and $\triangle LRK$, so I need $IC \cap LR = Q$, $IJ \cap LK = A$ and $CJ \cap RK = P$ to be collinear, i.e. $Q ∈ AD$, though I myself am somewhat overwhelmed at this point. Does anyone know of a solution that doesn't rely heavily on analytic methods?

 A: Some introductory words first, no formulas, no pictures yet, words regarding my process of searching and finding a solution. Then why this solution should be considered still as beautiful. Then some further comments on first tries to solve the problem inside projective geometry, and why this failed for me. Please skip, if it feels annoying at some (early) point.
The key to the solution was indeed to exhibit the (equivalent) situation using an economical statement (few points, and only points that determine the geometric constellation). As given, we have constructions inside the triangles $\Delta ABC$, so that $I,J$ appear, and $\Delta ACD$, so that $K,L$ appear, then we build the lines $IK$ and $JL$, and their intersection should be on the angle bisector in $C$ in $\Delta BCD$. After drawing some pictures, and trying to see which are the (independent) parameters determining the configuration of points, the problem becomes simpler.
First of all, from the whole triangle $\Delta ABC$ we need only $I,J$.
We fix $A,C$, we need them.
But we remove $B$ as starting point from the story, consider the point $I$ only. Then $J\in AI$ is obtained by constructing a right angle in $C$. Similarly, $D$ is not an imperative, we need only $K$, then construct $L\in AK$ so that we have again a right angle in $C$ in the corresponding triangle $\Delta KCL$. A lot of angle bisectors are now eliminated from the picture.
To have more symmetry in the statement, i will keep $I,J$ in the notation, and use $I',J'$ instead of $K,L$. Now the picture is determined by the angles in $A,C$ in the two triangles $T=\Delta AIC$ and $T'=\Delta AI'C$. With this simpler constellation of points, it is easier to isolate a situation to show a concurrence of lines (or alternatively to intersect two lines, and show a colinearity of three points). We finaly show a Ceva / Menelaus shaped relation. The involved proportions have to be expressed in terms of the free parameters of the problem. They are the angles in $T,T'$. Well, the sine theorem is handy in such situations, and some trigonometric expressions do occur. Arguably, this destroys the geometry. Most times, people that are not typing solutions come with this argument. My point of view is that if the involved $\sin$-relations are simple, a posteriori a purist user may easily find her or his road to rewrite. So there will be no complicated trigonometric yoga below.
One more comment. My first try was to use projective relations. For instance the fact that on the line $AI$ the points $A$, $A_B:=AI\cap BC$, $I$, $J$ are building a harmonic tuple. Similarly, the points $A$, $A_D:=AK\cap CD$, $K$, $L$ are building a harmonic tuple. Then using for instance projectivity, the lines $IL$, $JK$, $A_BA_D$ are concurrent in a point, call it $\Omega$. Now one can try to find the position of $\Omega$ on the segment $A_BA_D$, show it is on the angle bisector in $C$ in $\Delta BCD$. But there is a problem regarding the beauty now again. We have to use either $IL$ or $JK$, and we need to use angles, formulas get immediately angry while writing them, still no problem, but there is no clear (computational) idea to show. Nevertheless, some properties involving the points $A_B,A_D$ may be mentioned as bonus to complete the picture. So let us start.

We remove from the picture all unneeded letters, use $I',J'$ instead of $K,L$ to keep the symmetry in the picture, and restate equivalently:
Proposition 1: On the segment $AC$ consider in different half-planes the triangles $T=\Delta AIC$, $T'=\Delta AI'C$. We denote by $a,a'$ the two angles in $A$ in $T,T'$. And by  $c,c'$ the two angles in $C$ in $T,T'$. Construct now $J\in AI$ and $J'\in AI'$ so that the triangles $\Delta ICJ$, $\Delta I'CJ'$, have each a right angle in $C$. Draw the angle bisector of $\widehat {ICI'}$ and reflect $A$ into $A^*$ w.r.t. to it. So  $\widehat{ACI}$ and $\widehat{A^*CI'}$ have the same measure (with opposite sign). (This property can be used to construct $A^*$ alternatively.)
Then: The lines $IJ'$, $I'J$, $CA^*$ are concurrent.

Proof: Define the point $\Omega$ as $\Omega:=IJ'\cap CA^*$. We will show $\Omega\in I'J$. A good and simple way to show this is by checking the formula of Menelaus in $\Delta AIJ'$ w.r.t. the to-be-secant-line passing through $\Omega\in IJ'$, $I'\in J'A$, and $J\in AI$:
$$
\tag{$*$}
1\overset?=
\frac{\Omega I}{\Omega J'}\cdot
\frac{I'J'}{I'A}\cdot
\frac{JA}{JI}\ .
$$
So we start computing the R.H.S. from above.
$$
\begin{aligned}
\frac{\Omega I}{\Omega J'}\cdot
\frac{I'J'}{I'A}\cdot
\frac{JA}{JI}
& = 
\frac{\Omega I}{CI}\cdot
\frac{C I}{C J'}\cdot
\frac{CJ'}{\Omega J'}\ \cdot\ 
\frac{I'J'}{I'A}\ \cdot\ 
\frac{JA}{JI}
\\
& = 
\frac{\sin c'}{\sin\hat\Omega_1}\cdot
\frac{C I}{C J'}\cdot
\frac{\sin\hat\Omega_2}{\sin (90^\circ + c)}\ \cdot\ 
\frac{I'J'}{I'A}\ \cdot\ 
\frac{JA}{JI}
\\
&
\qquad\qquad\text{($\hat\Omega_1$, $\hat\Omega_2$ are the two angles between $C\Omega$ and $IJ'$)}
\\
& = 
\frac{\sin c'}{\sin (90^\circ + c)}\cdot
\frac{C I}{JI}\cdot
\frac{I'J'}{C J'} \cdot 
\frac{JA}{CA}\cdot 
\frac{CA}{I'A}
\\
& = 
\frac{\sin c'}{\sin (90^\circ + c)}\cdot
\frac{\cos\widehat{CIJ}}{\sin\widehat{CI'J'}}\cdot
\frac{\sin (90^\circ + c)}{\sin (90^\circ -c-a)} \cdot 
\frac{\sin(a'+c')}{\sin c'}
\\
&=1
\ .
\end{aligned}
$$
Here, $\widehat{CIJ}=a+c$, and $\widehat{CI'J'}=a'+c'$, as exterior angles for $T,T'$ in $I,I'$ respectively. So the question mark in $(*)$ is elucidated.
$\square$


We are done, but to have the beauty back, we can add some bonus ingredients and conclude in the given situation:
Proposition 2 (Bonus): Let $ABCD$ be a quadrilateral. In $\Delta ABC$ we construct $I,J, A_B$ as the incenter, the $A$-excenter, and the intersection of the $A$-angle bisector with the opposite side. In $\Delta ACD$ we construct in similar manner $K,L,A_D$.
Then $IL$, $JK$, $A_BA_D$, and the $B$-angle bisector of $\Delta BCD$ are concurrent in a point, call it $\Omega$.
And $IK$, $JL$, $A_BA_D$, and the $B$-angle exterior bisector of $\Delta BCD$ are concurrent in a point, call it $\Xi$.
And the $4$-tuples $(AA_B;IJ)$, $(AA_D;KL)$, $(A_BA_D;\Omega\Xi)$ are harmonic. (The associated cross ratios are $-1$.)

Proof: Using Proposition 1, we know $\Omega:=IL\cap JK$ lies on the $B$-angle bisector. Then in $\Delta CAA_B$ the $C$-angle bisectors (internal and external) are marking on $AA_B$ a harmonic ratio.
Same for $\Delta CAA_D$.
Denoting the cross ratio of four points by square brackets, we have thus
$$-1 = [AA_B;IJ]= [AA_D;KL]\ .$$
Recall that the cross ratio is invariated when projecting from a point from a line onto an other line. A converse property is also true, so from
$[AA_B;IJ]=-1=[AA_D;LK]$ we obtain the concurrence of the three lines
$A_BA_D$, $IL$, $JK$. (Indeed, project from $\Omega:=IL\cap JK$ the line $AI$ onto the line $AJ$. Then $A\to A$, $I\to L$, $J\to K$, and $A_B$ goes to a point which conserves the cross ratio, this point is unique, so it is $A_D$, so $\Omega\in A_BA_D$.)
Similarly, using $[AA_B;IJ]=-1=[AA_D;KL]$ we get the well defined point $\Xi$ from the statement. Applying Proposition 1. for $A,C,I,I'=L$ instead of $A,C,I,I'=K$
we get $\Xi$ instead of $\Omega$,
we use the angles $a,a';c, c'+90^\circ$ instead of $a,a';c,c'$, so the point $A^*$ gets a $90^\circ$-rotation around $C$, so the new angle bisector $CA^*$ is the exterior angle bisector. We are done.
$\square$
A: Hints:
Here are  two strategies you can try:
1):
-Draw circumcircle of triangle BCD.
-Draw a circle on center J tangents to sides( or their extension) of triangle ABC. You have to prove following points:
1- IL is perpendicular on KJ.
2- Extend CM to touch the circucircle of triangle BCD at point G. Connect J to G. JG is perpendicular bisector of BD, therefore CM is bisector of $\angle BCD$.
2):
1-Draw a perpendicular on CM from point C to touch the circle centered on J and tangent on sides of triangle ABC at point P.
2- Connect C to J, it intersect the circle at C'.  Extend it to meet the circle again at H.
3- Mark tangent point of this circle with side BC as N. Connect N to J and extend it to touch the circle at E. Arcs C'N and EH are equal, therefore arcs NP and PH are equal, that is CP is bisector of angle $\angle BCJ$ which in turn means CM is bisector of $\angle BCD$.
A: Lemma. Let $AP$, $AQ$ and $AS$, $AT$ be two pairs of isogonal lines with respect to $\angle BAC$. Let $PS \cap QT = X$ and $PT \cap QS = Y$. Then, $AX$, $AY$ are isogonal lines with respect to $\angle BAC$.
Drawing inspiration from other answers, it occurred to me that isogonality of lines played a rather important role, and hence provided motivation for the application of the above lemma, referred to as the Isogonal Lines Lemma, a proof of which can be found in the book.
With this in mind, we proceed with the solution to the problem.
Proof. Let $IL \cap JK = M$, we shall show that $CM$ bisects $\angle BCD$.

Note that $\angle ICJ = \angle KCL = 90°$, so $CJ$ and $CL$ are isogonal with respect to $\angle ICK$. By the Isogonal Lines Lemma applied to lines $CI$ and $CK$ along with $CJ$ and $CL$, we see that the lines joining $C$ to $IJ \cap KL = A$ and to $IL \cap JK = M$ are isogonal as well, so $\angle ACI = \angle MCK$. Consequently,
$\begin{align*}
\angle BCM = \angle BCA + \angle ACM &= 2\cdot \angle ACI + \angle ACM \\
&= \angle MCK + (\angle MCK + \angle ACM) \\
&= \angle MCK + \angle ACK \\
&= \angle MCK + \angle DCK \\
&= \angle DCM,
\end{align*}$
as desired.
