lemma:In triangle ABC,M is the midpoint of BC so we have (cot ∠CAM=1/2(cot∠C-cot∠B)) So if ∠AND=∠DNC we have cot∠ADB-cot∠ABD=cot∠CDB-cot∠CBD we should prove ∠DLC=∠BLC or cot∠DLC=cot∠BLC We know cot∠DLC=cot∠ACD-cot∠DAC=cot∠ABD-cot∠DBC also cot∠BLC =cot∠BCA-cot∠BAC=cot∠BDA-cot∠BDC and we know cot∠ADB-cot∠ABD=cot∠CDB-cot∠CBD thus cot∠DLC=cot∠BLC and ∠DLC=∠BLC so AC is the bisector of the angle BLD.

If I would have enough information about Pole and Polar, my solution would be shorter. Anyway... Let $O$ be the circumcentre of $ABCD$. Claim: The tangents to $A$ and $C$, and the line $BD$ are concurrent. Proof: Let $E$ be the reflection of $C$ over $ON$. $CN=NE$ and $ON\perp CE$. Since $BD\perp ON$, we have $BD\parallel CE$ and $\angle BNC=\angle NCE=\angle NEC = \angle ANB$. So $A,N,E$ are linear. $\angle ANC = \angle AOC = 2\cdot AEC$ means $A,N,O,C$ are cyclic. Let $P$ be the point where $OL$ meets $(ANOC)$. Measure of smaller arc $AP$ is $\angle AOP/2$, which is equal to $\angle ANB/2$. So $NB$ cuts $(ANOC)$ at $P$. Since $\angle PNO = 90^\circ=\angle PAO$, $PA$ is tangent to $(ABCD)$. We have $PA=PC$, so $PC$ is also tangent to the circle. $\blacksquare$ Now the problem asks for $AC$ is the angle bisector of $\angle BLD$. Since $\angle ALP = 90^\circ$, $\angle BLP = 90^\circ - \angle BLA$. If $AC$ is angle bisector of $\triangle BLD$, then $LP$ should be external angle bisector of $\triangle BLD$. This is just brainstorming. If so, Claim: $\frac{PB}{PD} = \frac {BF}{FD}$ Proof: Let $PD\cap AC = \{F\}$. The power of $F$ with respect to $(ABCD)$, $BF\cdot FD = AF \cdot FC$. The power of $P$ with respect to $(ABCD)$, $PA^2 = PB \cdot PD$. From Stewart in $\triangle PAC$, $PF^2 = PA^2 - AF\cdot FC$. Merging above equations gives \[\begin{array}{rcl}PF^2 &=& PB\cdot PD - BF\cdot FD \\ &\Rightarrow & (PD-FD)^2 = PB\cdot PD - BF\cdot FD \\ &\Rightarrow & PD^2 + FD^2 - 2\cdot PD\cdot FD = PD^2 - FD\cdot PD - BF\cdot PD - BF\cdot FD \\ &\Rightarrow & BF\cdot PD = -FD^2 + FD\cdot PD - FD\cdot BF = FD(PD-BF-FD)=FD\cdot PB \end{array}\] Convert the last one into a ratio \[\frac {BP}{PD} = \frac{BF}{FD} \blacksquare\] We have all ingredients for internal and external angle bisectors. We have $\frac {BP}{PD} = \frac{BF}{FD}$ The angle between internal and external bisector is $\angle PLF = 90^\circ$ Let's use technical words. The ratios of distances of $P$ and $F$ to $B$ and $D$ are equal. Locus of such points is an Apollonius circle of $B$ and $D$. $PF$ is the diameter of this circle. Since $\angle PLF = 90^\circ$, $L$ is also a point on locus. So \[\frac {BP}{PD} = \frac{BF}{FD} = \frac{BL}{LD}.\] This yields, $LF$ is an angle bisector of $\triangle BLD$ $\blacksquare$.

Sorry for a revive, but a much simpler solution.. Suppose $ AC \cap BD=E $, the perpendicular bisector of $ AC $ meets $ BD $ at $ P $ and the perpendicular bisector of $ BD $ meets $ AC $ at $ Q $. Since $ P $ is the intersection of the angle bisector of $ \angle ANC $ and perpendicular bisector of $ AC $, then $ ANCP $ is cyclic $ \implies EN.EP=EA.EC=EB.ED $. From cyclic $ PQNL $ we get $ EP.EN=EQ.EL \implies EQ.EL=EB.ED \implies BLDQ $ is cyclic $ \implies Q $ is the midpoint of arc $ BQD $ of circle $ (BLDQ) \implies LQ $, or $ AC $ is the angle bisector of $ \angle BLD $.

Let $J, K$ be the second intersections of $AN, CN$ respectively with the circumcircle of $ABCD$. Clearly $AK\parallel CJ\parallel BD$ and, consequently $\angle AND=\angle ABC\therefore\triangle AND\sim\triangle ABC\implies\frac{ND}{BC}=\frac{AD}{AC}$, hence $ND\cdot AC=BC\cdot AD$. In a similar way $BN\cdot AC=AB\cdot CD$; with $BN=ND$ we get $AB\cdot CD=AD\cdot BC$, or $ABCD$ is a harmonic quadrilateral, but the given problem is a well known property of harmonic quadrilaterals. Best regards, sunken rock