Prove that $IL,JK$ and angle bisector of angle $BCD$ are concurrent Given a convex quadrilateral $ABCD$. In $\Delta ABC$, $I$ is the incentre and $J$ is the excentre opposite to vertex $A$. Similarly, $K$ is the incentre and $L$ is the excentre opposite to vertex $A$ of $\Delta ACD$. Prove that the three lines $IL,JK$ and the angle bisector of angle $BCD$ are concurrent.
I tried assuming some angles in one if the triangle. Used some trigonometry, but the problem just got nasty. I don't think pure geometry would do it. The problem looks good. If possible, please solve using trigonometry and/or algebraic geometry. Thanks.
 A: With somewhat-altered labels, we have this figure for $\square ABOC$ with incenters $P$ and $Q$ (and inradii $p$ and $q$) and excenters $R$ and $S$ (and exradii $r$ and $s$).

We embed the figure in the coordinate plane with $O$ at the origin, and:
$$A = a\;(1,0) \qquad B = b\;(\cos 2\beta,\sin 2\beta) \qquad C = c\;(\cos 2\gamma,-\sin 2\gamma)$$
$$\begin{align}
P &= \frac{p}{\sin\beta}(\phantom{-}\cos\beta,\sin\beta) \qquad 
Q = \frac{q}{\sin\gamma}(\phantom{-}\cos\gamma,-\sin\gamma) \\[4pt]
R &= \frac{r}{\cos\beta}(-\sin\beta,\cos\beta) \qquad
S = \frac{s}{\cos\gamma}(-\sin\gamma,-\cos\gamma)
\end{align}$$
In/exradius formulas are known, and we can compute, for instance,
$$\frac{p}{\sin\beta} = \frac{|\triangle OAB|}{\sin\beta\;(a+b+e)/2} = \frac{a b \sin 2\beta}{\sin\beta\;(a+b+e)} = \frac{2 a b\cos\beta}{a+b+e}$$
Likewise,
$$\frac{q}{\sin\gamma} = \frac{2 a c\cos\gamma}{a+c+f} \qquad
\frac{r}{\cos\beta} = \frac{2 a b \sin\beta}{a-b+e} \qquad
\frac{s}{\cos\gamma} = \frac{2 a c \sin\gamma}{a-c+f}$$
From here, brute-force symbol manipulation (and, very likely, a slick geometric argument that has eluded me) determines the intersection point, $K$, of lines $\overleftrightarrow{PS}$ and $\overleftrightarrow{QR}$:
$$K = \frac{a b c}{a b + a c + c e + b f }\; \left(\;\cos 2 \beta + \cos 2 \gamma\;,\;\sin 2\beta - \sin 2\gamma\;\right) = \frac{a\;( c B + b C )}{a b + a c + c e + b f}$$
Since $c B + b C$ is a direction vector for the bisector of $\angle BOC$, we have that $K$ is on this bisector and is therefore the sought-after point of concurrency. $\square$

Edit. We can reduce some symbol manipulation, and get an interesting general concurrence criterion, by deferring the radius calculations. For instance, with
$$\begin{align}
P = p^\prime\;(\phantom{-}\cos\beta,\sin\beta) &\qquad Q = q^\prime \;(\phantom{-}\cos\gamma,-\sin\gamma) \\ 
R = r^\prime\;(-\sin\beta,\cos\beta) &\qquad S = s^\prime\;(-\sin\gamma,-\cos\gamma)
\end{align}$$
(where $p^\prime$, etc, are distances from $O$ along various angle bisectors, although not necessarily related to any in/exradii) we find that
$$K = \frac{\cos(\beta+\gamma)}{p^\prime r^\prime + q^\prime s^\prime + (p^\prime q^\prime + r^\prime s^\prime ) \sin(\beta + \gamma)}\;\left(\;q^\prime r^\prime (P-S) + p^\prime s^\prime (Q - R) \;\right)$$
Ignoring the multiplied scalar, we force $K$ onto the bisector of $\angle BOC$ by substituting its components into the equation $y = x\,\tan(\beta-\gamma)$. The result is this relation:
$$q^\prime r^\prime s^\prime \cos\beta - p^\prime r^\prime s^\prime \cos\gamma - p^\prime q^\prime s^\prime \sin\beta + p^\prime q^\prime r^\prime \sin\gamma = 0$$
Or, better:

$$\frac{\cos\beta}{p^\prime} - \frac{\sin\beta}{r^\prime} = \frac{\cos\gamma}{q^\prime} - \frac{\sin\gamma}{s^\prime} \tag{$\star$}$$

(This seems like something that should have an elegant geometric derivation.) Substituting $p^\prime = 2 a b \cos\beta/(a+b+e)$, etc, causes each side of $(\star)$ to reduce to $1/a$.
