angle chasing in quadrilaterals 
Let $ABCD$ be a convex quadrilateral 
with 
  $\measuredangle{ABD} = 18^{\circ}$, $\measuredangle{ACB} = 54^{\circ}$,
  $\measuredangle{ACD} = 36^{\circ}$ and 
  $\measuredangle{ADB} = 27^{\circ}$.
Let $\{P\}$
  be the intersection of the two diagonals $AC$ and $BD$. 
What is the degree value of $\measuredangle{APB}$?


 A: After some angle chasing, we have the following:-

The configuration seems like C is the circum-center of the circle DAB.
Added:-
The tricky part is:- "Is C really the circum-center of $\triangle ABD$?"

This is done by extending AC to cut the circum-circle at F. Again by angle chasing, we get $\beta = \angle 1 = \angle 2$. This means $\triangle CFB$ is isosceles. The same is true for $\triangle CFD$. Then, point C (not the red dot anymore) is indeed the circum-center of $\triangle ABD$ (because (1) CB = CF = CD and (2) B, F, D are points on the circumference of the same circle).
Finally, we have $\angle CDB = \angle CBD = 45^0$. Result follows.
A: Let $CK$ be a bisector of $\Delta ADC$ and $CL$ be a bisector of $\Delta ABC$.
Thus, $KCLA$ is cyclic and from here $\measuredangle KLA=\measuredangle KCA=18^{\circ}=\measuredangle ABD$.
Thus, $KL||DB$, which says $AC:DC=AK:KD=AL:LB=AC:BC$,
which gives $DC=BC$, $\measuredangle CBA=45^{\circ}$ and $\measuredangle APB=54^{\circ}+45^{\circ}=99^{\circ}.$
A: Let S be the center of the circumcircle of triangle ABD. D is a point on that circumcircle, and ∠ADB = 18∘, so therefore, ∠ASD = 36∘. Consider the circumcircle of ASD. S is a point on that circumcircle and ∠ASD = 36∘, so therefore for each point on that circumcircle Q and only for those points, ∠AQD = 36∘ is a true statement.
We know ∠ACD = 36∘, so therefore, C is on the circumcircle of ADS.
Analogically, since ∠BCA = ∠BSA = 2 * ∠BDA = 54∘, C is also on the circumcircle of ABS. 
Out of this, we can conclude points C and S are on the intersections of ABS's and ADS's circumcircles. The only such intersections are points A and C. Obviously, S is not the same point as A (otherwise, ∠BDA = ∠DBA would be true), and neither is C (otherwise, ∠ACB = 54∘ makes no sense) so we conclude S is C. 
Therefore, C is ABD's circumcenter. Out of that, we can get DC = BC (two radii in the same circle), and by that we conclude BCD is bilateral and ∠DBC = ∠BDC. 
We know ∠BCD = 36∘ + 54∘ = 90∘. Therefore , ∠DBC=45∘.
Finally, ∠APB = 54∘ + 45∘ = 99∘.
