First,some quick notations.Let $Q=IL \cap JK$.Let $AB=a,BC=b,CD=c,DA=d,AC=e,\angle ACB=B,\angle ACD=D$.Also denote by $r_{ABC}$ the inradius of $ABC$,by $r_{eACD}$ the exradius corresponding to $A$ of $ACD$,by $d(W,XY)$ the distance from $W$ to $XY$,and by $S_x$ the area of figure $x$. We apply Menelaus' on $J-Q-K$ and triangle $AIL$ to get that $\frac{AJ}{IJ}\frac{IQ}{LQ}\frac{LK}{AK}=1$.By Sine Law we get $AI=\frac{1}{\sin{(BAC/2)}}\frac{a+e-b}{2}$,$AJ=\frac{1}{\sin{(BAC/2)}}\frac{a+e+b}{2}$,$AK=\frac{1}{\sin{(DAC/2)}}\frac{d+e-c}{2}$, $AI=\frac{1}{\sin{(DAC/2)}}$$\frac{d+e+c}{2}$ and also $IJ=\frac{1}{\sin{(BAC/2)}}b,KL=\frac{1}{\sin{(DAC/2)}}c$. Quick calculations now give us $\frac{IQ}{LQ}=\frac{b(d+e-c)}{c(a+b+e)}$ and its "cousins" $\frac{IQ}{IL}=\frac{b(d+e-c)}{K}$ and $\frac{LQ}{IL}=\frac{c(a+b+e)}{K}$,where we denoted $K=ac+ec+bd+be$. What we need to prove is that $Q$ lies on $BCD$'s bisector;this is done if we prove $d(Q,BC)=d(Q,DC)$. Let's first calculate $d(Q,DC)$.We notice that $I$ and $L$ are on different sides of $DC$.Let's assume WLOG $X$ is on the same side as $I$.Thus we can find that $d(Q,DC)=\frac{LQ}{LI}d(I,DC)-\frac{IQ}{IL}d(L,DC)$$(*)$.Now by Sine Law again we obtain $d(L,DC)=r_{eADC}$ and $d(I,DC)=r_{ABC}*\frac{sin{(\frac{B}{2}+D)}}{sin{(\frac{B}{2})}}$.Now we couple these with the well-known formulas $S_{ABC}=\frac{a+b+e}{2}r_{ABC}$,$S_{ACD}=\frac{d+e-c}{2}r_{eACD}$ and mash our formula $(*)$ by easy computations into: $d(Q,DC)=\frac{1}{K}(2cS_{ABC}\frac{sin{(\frac{B}{2}+D)}}{sin{(\frac{B}{2})}}-2bS_{ADC})$. Now on to $d(Q,BC)$;this time,we do not know for sure whether $I$ and $L$ lie on the same side of $BC$;but we can still assume $WLOG$ that they do;now obviously $Q$ lies on the same side as both of them,so the formula for the distance is: $d(Q,BC)=\frac{LQ}{LI}d(I,BC)+\frac{IQ}{IL}d(L,BC)$. By once again mashing and bashing by known formulae we arrive at $d(Q,BC)=\frac{1}{K}(2cS_{ABC}+2bS_{ADC}\frac{cos{(\frac{D}{2}+B)}}{cos{(\frac{D}{2})}})$ Now that we have the formulae for both distances,proving that they are equal is reduced to a neatly-looking $2cS_{ABC}(1-\frac{sin{(\frac{B}{2}+D)}}{sin{(\frac{B}{2})}})=-2bS_{ACD}(1+\frac{cos{(\frac{D}{2}+B)}}{cos{(\frac{D}{2})}})$. By the area formula,$S_{ABC}=be\sin{B},S_{ADC}=ce\sin{D}$ so after some simple simplification we are left to prove just that $cos{\frac{B}{2}}(sin{(\frac{B}{2})}-sin{(\frac{B}{2}+D)})=-sin{\frac{D}{2}}(cos{(\frac{D}{2})}-cos{(\frac{D}{2}+B)})$ which is standard trig manipulation,$QED$.

My solution : Let $ X=IL \cap JK $ and $ \ell, \tau $ be the bisector of $ \angle DCB, \angle (\ell, CA) $ , respectively . Since $ \angle (CI, \tau)=\angle (CK, \tau) $ and $ \angle (CJ, \tau)=\angle (CL, \tau) $ , so we get $ (CI, CJ; \ell, CA)=(CK, CL; CA, \ell) $ . ... $ (\star) $ From the Dual of Desargue Involution theorem $ \Longrightarrow (CI, CJ; CX, CA)=(CK, CL; CA, CX) $ , so combine $ (\star) $ we get $ CX \equiv \ell \Longrightarrow X \in \ell $ . i.e. $ IL, JK $ and the bisector of $ \angle DCB $ are concurrent Q.E.D

Note: the property is actually valid for any 4 points in the plane,not just for a convex quadrilateral.

My solution. Lemma. Given 4 angles $\alpha, \alpha’, \beta, \beta’$ such that $\alpha+\beta = \alpha’+\beta’ <180$ and $\frac{sin\alpha}{sin\beta}$ = $\frac{sin\alpha’}{sin\beta’}$. We’ll have $\alpha = \alpha’, \beta = \beta’$ Proof. Construct $\triangle{CAB}, \triangle{C’A’B’}$ such that $\angle{A}=\alpha, \angle{B}=\beta, \angle{A’}=\alpha’, \angle{B’}=\beta’$ $\Rightarrow$ $\angle{C}=\angle{C’}$ (because $\alpha+\beta = \alpha’+\beta’$) and $\frac{CB}{CA} = \frac{CB’}{CA’}$ $\Rightarrow$ $\triangle{ABC}$$\sim$$\triangle{A’B’C’}$ and the conclusion follows

Back to our main problem. $IL\cap{JK} = S$ Apply the Ceva theorem, we have: $\frac{sin\angle{SCK}}{sin\angle{SCI}} = \frac{sin\angle{SKC}}{sin\angle{SKI}}.\frac{sin\angle{SIK}}{sin\angle{SIC}} = \frac{sin\angle{JKC}}{sin\angle{JKI}}.\frac{sin\angle{LIK}}{sin\angle{LIC}} = (\frac{sin\angle{JCK}}{sin\angle{JCI}}.\frac{sin\angle{JIC}}{sin\angle{JIK}}).(\frac{sin\angle{LKI}}{sin\angle{LKC}}.\frac{sin\angle{LCK}}{sin\angle{LCI}}) = \frac{sin\angle{JIC}}{sin\angle{JIK}}.\frac{sin\angle{LKI}}{sin\angle{LKC}} = \frac{sin\angle{AIC}}{sin\angle{AIK}}.\frac{sin\angle{AKI}}{sin\angle{AKC}} = \frac{sin\angle{ACI}}{sin\angle{ACK}}$ Now apply the lemma with the notice that $\angle{SCK}+\angle{SCI} = \angle{ACI}+\angle{ACK}$, we receive: $\angle{SCK} = \angle{ACI}, \angle{SCI} = \angle{ACK}$ (1) On the other hand: $CI, CK$ are the angle bisectors of $\angle{ACB}, \angle{ACD}$ (2) (1), (2) $\Rightarrow$ $CS$ is the angle bisector of $\angle{BCD}$ Q.E.D

Attachments:

tranquanghuy7198 wrote: \[(\frac{sin\angle{JCK}}{sin\angle{JCI}}.\frac{sin\angle{JIC}}{sin\angle{JIK}}).(\frac{sin\angle{LKI}}{sin\angle{LKC}}.\frac{sin\angle{LCK}}{sin\angle{LCI}}) = \frac{sin\angle{JIC}}{sin\angle{JIK}}.\frac{sin\angle{LKI}}{sin\angle{LKC}}\] How can you say that?? Where are other two fractions??

Actually this is very easy using harmonic divisions Let $IJ\cap BC=E,KL\cap CD=F$.since cross ratios $[AIEJ],[ALFK]$ are same (harmonic) we have $IL,EF,JK$ are concurrent say at $X$. Next we have that $IF,EKAX$ are concurrent.Indeed this follows as pencil$[IA,IK,IF,IX]$ is harmonic Now apply ceva in $AEF$ and get $EX/XF=EC/CF$ so we are done.

India Postals 2015 Set 2 wrote: Let $ABCD$ be a convex quadrilateral. In the triangle $ABC$ let $I$ and $J$ be the incenter and the excenter opposite to vertex $A$, respectively. In the triangle $ACD$ let $K$ and $L$ be the incenter and the excenter opposite to vertex $A$, respectively. Show that the lines $IL$ and $JK$, and the bisector of the angle $BCD$ are concurrent. Let $X=IL\cap JK$, since $\angle JCI=\angle JCK=90^\circ\implies CJ, CL$ are isogonal w.r.t $\angle ICK$, using Isogonal Line Lemma we have $CA, CX$ are isogonal w.r.t $\angle ICK\implies \angle ACI=\angle XCK$. $$\angle XCB=2\angle ACI+\angle ACS=\angle KCD+\angle KCS=\angle XCD.\blacksquare$$