First, let's finish filling in some information. If angle AED is 70, then angle BEC must also be 70. Since their sum is 140, the leftover degrees must add up to 220. Since angles AEB and DEC are equal, that means they are each 110 degrees. Angles EDC and ECD must also be equal, so they are both (180 - 110)/2 or 35 degrees. Since ADE and EDC are complimentary, angle ADE must be 55 degrees. Angle EAD is also 55 degrees. Now that we've finished that, we can begin solving for AE.
Seperate this into two triangles, bisected by line segment AC. The angles for this triangle are EAD (55 degrees), DCE (35 degrees), and ADC (90 degrees). If DC is 15, we can solve for AC by taking the 15/cos(35). AE is one half of that, so it must be B.
14. We already determined this was D.
15. We already determined this was C.
16. This one must be 90 degrees.
In question 13, the diagonals of a rectangle bisect each other, so it's pretty obvious what AE is.
In question 14, the angles at the corners measure 90 degrees. So the adjacent angle to 55 degrees must be 35 degrees. The supplementary angle to 70 degrees is 110 degrees, so you have a triangle with angles 35, 110 and you can figure out the third. (gotta total 180, you know)''
In question 15, triangle AED is isosceles with a vertex angle of 70 degrees. That leaves 110 degrees to be divided equally among the base angles.
In question 16, a rectangle has 4 right angles.