I just solved the first one. Not that difficult. You draw two parallel lines to line AB that intersect E and D. This creates a regular trapezoid. The diagonals are congruent. I don't feel like writing out the rest of the steps, but here is a link to the scanned image of it.
I disagree. I got 20 (you can check it with a protractor if the figure is drawn to scale). Triangle ADE is not isosceles. The diagonals within the trapezoids are not parallel. The lower diagonal (BD) makes an angle of 60 degrees with AB; the upper diagonal (DE) is at 50 degrees from a line running through D, parallel to AB.
Look at my work before you mod me down. The lines I drew were not drawn with a ruler. If x is 10, then DE is 60 degrees from the line running through D (not 50), parallel to AB, just like the angle DBA.
If a protractor says one thing, and the math using known values says another, I'm going to trust the math.
I see where according to the figure, triangle ADE must be isosceles, but I don't trust that the figure is actually drawn to scale. All the measurements of angles I got by calculations. Can you find a flaw in one of those?
Check again. The figure is to scale. I worked it out first and then checked with a protractor. Your assumption that the diagonals within the trapezoids are parallel is incorrect. As a result, angles that you have assigned to the triangles in the upper trapezoid are also incorrect. The fact that triangle ADE is not isosceles makes it impossible for x to be 10.
> Your assumption that the diagonals within the trapezoids are parallel is incorrect.
Do you mean his assumption that the diagonals within the trapezoids are congruent is incorrect? I'd be curious to see your diagram or proof to see where you differ. Obviously, it makes a lot more sense that the diagram is drawn to scale.
I got x=25. I didn't draw any new lines.. I just worked out all the angles I could within the diagram, and then played x and y (other unknown angle within small triangle containing x) off against one another using dead-simple algebra.
Along the way I got:
y+x=130
x=y-80
Another hint - you know that point E has three angles - 30, x, (180-30-x). Similar logic works on the other side too.
answer is 20.
parallel line to line AB in D point gives us F point, after some work we get EF=FD and angle FED=50, but angle AEB=30.
So, here is the result angle DEA = FEB - AEB = 20 : )
The first one isn't actually solvable; you get a system of equations with infinite solutions. The second hint for it implies by looking at a drawing to scale and putting in a few extra lines you'll get it, but it sounds more like you're eyeballing angles. You might as well just pull out the protractor.
