In the diagram below, D is a point on segment CE such that segments AD and BE are parellel and point A is not on segment BC. Line segments AD and BC intersect at point P.
If angle CAD = 67 degrees, then what is angle CBE?
The answer is 134.
First of all, the question never specified that ACD was not 90 degrees, and without it there would still be sufficient info, but this just makes it easier to visualize things (You can prove that CP = PA without this assumption). Angle CAP is 67 degrees, and let angle PBA be x degrees. We can also construct it so that CD is parallel to AB, and CA makes a 90 degrees angle to both lines. Thus, angle PAB is 90 - 67 = 23 degrees. Since E is collinear with CD, then CE is parallel to AB, and DE is parallel to AB, so shape ABED makes a parallelogram, thus angle ADE is 180 - 23 = 157 degrees. Angle ABE is also 157 degrees, so angle CBE is 157 - angle PBA = 157 - x degrees. Angle CDP is 180 - 157 = 23 degrees, and angle PCD is 90 - 67 = 23 degrees, thus CP = PD, and angle CPD is 180 - 23 - 23 = 134 degrees. By corresponding angles, angle CBE = angle CPD = 157 - x = 134 degrees, and because we are looking for angle CBE, angle CBE = 134 degrees (you dont even really need the x, but its more visuals).