how is this problem possible? how can this be when there is no X?
There certainly is a point "X"! You are told that X lies on the line BD, extended beyound B. We are not told how far from B X is but that is not important because all we are concerned with is the angle. If the angle "ABX" is 110 degrees and since sides BX of angle ABX and BD of angle ABD lie on the same line, the angles are supplementary so that angle "ABD" is 180- 110= 70 degrees.
I'm not sure where I should go from here. If I were given this problem, I would ignore most of the information and immediately note that because angle A is 50 degrees, it must cut a 100 degree arc off the circle. But that means the other part of the circle is a 360- 100= 260 degrees and so angle C has measure 130 degrees. But because you are given that additional information you may not know or be able to use the fact that an angle, with vertex on the circle, cuts off an are of measure twice the degree measure of the angle.
I did consider using the fact that, because A and C are ends of a diameter, angle ABC is a right triangle. But that again depends upon the fact that half a circle has arc measure 360/2= 180 degrees and so the angle is 180/2= 90. That is, it derives from the "fact" I mentioned above. Do you know that a triangle that has all it points on a circle and one side a diameter is a right triangle?
(Be aware that we cannot assume that line AC bisects angle A, because we are not told that, so we cannot assume "symmetry" about that line.)
Your post is flawed.Angle A and angle ABX are related. If A=50 then ABX must be 115.If angle ABX is 110 then A=40.ADCB is a cyclic kite
Angle C is the supplement of A because opposite angles of inscribed quad are supplementry.
Above is wrong because it was not shown that AC is perpendicular to DB.
Angle A and C are supplementry because opposite angles of an inscribed quad are supplementry