Theorem : Any problem stated as "let ABC and P inside" or "let ABCD a quadrilateral" may be reduced to a "let ABC a triangle and D a point inside the circumcircle of ABC", eventually renaming the points. Any such problem in which all angles are multiples of 10 deg can then be overlayed on a regular 18-gon and its diagonals and is then equivallent to "in a regular 18-gon these such diagonals intersect in one point".
As in a regular 18-gon, at reflections and rotations, there are exactly 9 points which are non obvious intersections of three or more diagonals, conclusion :
There are at most 9 classes of "adventitious" quadrilaterals and point in triangles, with all angles multiples of 10 deg. Any problem in one class is equivallent to another problem of the same class, that is the proof of one is a one sentence proof : "because of the other one". And "at most" is very true from many equivallences Intersections X <=> some problem <=> intersection Y, as there are generally several ways of overlaying one problem on the 18-gon.
This dramatically decreases your list of problems to solve !
However in this specific case of today's problem, this is how it works to reduce this problem to one of the very first you already posed on 22 Oct. (see attachment)
We get at once that ABC = ACB = 40 deg, and that angle BPC = 180 - (30+40) - 40 = 70 deg Hence BPC is isosceles, and as angle BCP is a multiple of 20 deg, the figure can be overlayed on the 18-gon with C at center. This results into the proof being equivallent to "diagonals P5-P10 and P7-P16 intersect in Q on diagonal P2-P8" But P7-P16 being a diameter, this is equivallent to "diagonals P7-P16, P5-P10, P2-P8, P6-P12 and P4-P9 intersect in one point"
But now the old problem of Oct. 22 :
> ... Here is another problem that can be solved in the same way- > In an isosceles triangle A'B'C' with vertical angle 100 deg and base > angles 40 deg each, point P' is taken inside the triangle such that > angle P'B'C' = 30 deg and angle P'C'B' = 20 deg > Find angle P'A'C'. (In which I have subtituded all letters by ' to not confuse with today's problem)
is equivallent to "diagonals P2-P8 and P5-P10 intersect on P6-P12"
So this old problem is a proof of these 5 diagonals intersecting in one point, hence a proof of today's problem. Who ever has solved the proposed problem on Oct. 22 has then solved "at once" today's problem.
The solution of Oct. 22 was not given, for it was a "bonus" problem. Here it is (second figure in the attachment) :
Construct equilateral triangle B'C'D', A'D' is the perpendicular bisector of B'C' in both isosceles triangles. Hence angle A'D'C' = (60/2) = 30 = given P'B'C' Angle A'C'D' = 60 - 40 = 20 deg = P'C'B' as B'C' = C'D', triangles P'B'C' and A'D'C' are congruent. Hence A'C' = P'C' and triangle A'C'P' is isosceles, hence base angles, that is angle P'A'C' = 80 deg
And then : Hence these diagonals are concurrent, hence angle CPQ in today's problem is 30 deg.
Please also note that today's assignment may be misunderstood, for "angle ABD = 30 deg" may be in any of two directions, as well as ACD, resulting into 3 different cases in which "D is outside" (two last figures in attachement in addition to the "expected" one)