Have a go at this question ... it was from AIMO 2007 Intermediate Paper.
A point P is marked inside a regular hexagon ABCDEF so that angle BAP = angle DCP = 50 degrees. Find angle ABP.
In my opinion, its harder than it looks.
I'm sure there is a shorter solution to this problem. However, this long solution is the only solution I can come up with.
I isolated the quadrilateral ABCP. From the problem, BAP is 50 degrees. Since DCP is also 50 degrees, and since the interior angles of a regular hexagon are 120 degrees each, then BCP is 70 degrees. ABC is 120 degrees, and since the angles of a quadrilateral sum up to 360 degrees, then APC is also 120 degrees.
Triangle ABC is an isosceles triangle. AB = BC. Hence, BAC = BCA = 30 degrees.
That makes PAC = 20 degrees and PCA = 40 degrees.
Let's say AB = BC = 2. Using trigonometry (special triangles) on triangle ABC, AC = 2 * sqrt(3).
Using sine law on triangle APC, AP = 4 * (sin 40)
Using cosine law on triangle BAP, we find BP = 2. Hence, APB = 50 degrees and ABP = 80 degrees
AB = BP = BC, but I don't know how to make that conclusion without solving it the long way.