Explanation:
a reminder about a few things :
the sum of all angles in a triangle is always 180°.
the sum of all angles around a point on one side of a line is also always 180°.
similar shapes (like ABC and AEF) have the same angles.
they are similar, because all 3 sides from one triangle are parallel to their corresponding sides of the other triangle.
(b)
b + 5a - 40 = 180 (supplementary angles, b and 5a-40 are the complete set of angles around A on one side of BD).
b = 180 - 5a + 40 = 220 - 5a
80 + 2a + 220 - 5a = 180 (the sum of all angles)
-3a + 300 = 180
-3a = -120
3a = 120
a = 40
so, 2a = 2×40 = 80°, which is the same as the angle at F.
2 equal angles at both ends of a triangle side means that we have an isoceles triangle, and therefore A E = A F.
b = 220 - 5a = 220 - 5×40 = 220 - 200 = 20°
c = 2a = 80° (angles in similar shapes are the same).
d + 80 = 180
d = 100° (again supplementary angles this time around F)
e = 80 (again, angles in similar shapes are equal).
(c)
the angles between 2 intersecting lines are the same when one of these lines intersect with a line that is parallel to the other line.
because parallel lines simply imitate each other perfectly in all attributes except for the physical location. but slipped, angles are all the same.
also, the angles between 2 intersecting lines are the same on both sides of every one of the 2 lines, they are just left-right mirrored.
so,
x = 28°
as ED and EF are intersecting, CD (as parallel line to EF) intersects ED at the same angles.
28° is the angle on the "above" side of the intersection, and x is on the "under" side. because of the left-right mirroring between above and under sides, x = 28°.
x is part of the angle BDE, so the angle CDB is
160 - x = 160 - 28 = 132°
the angles CDB and ABD are together
132 + 48 = 180°
so, they are supplementary angles again, indicating they are again the sum of both angles around an intersection point of 2 lines (with the line BD). and that is only possible, if AB and CD are again parallel as before CD and EF.