ABCDE is a regular pentagon. A line is drawn from B to E. What is the size of angle ABE?

The internal angles of a regular pentagon are 108°.

(If you need proof I like to use external angles, rather than memorizing an internal angle formula: 360°/5 = 72°

=> internal angle = 180°-72° = 108°)

Drawing a line from B to E will form an obtuse Isosceles triangle with A forming the obtuse angle. Angle A is not affected, and remains 108°.

The sum of all the angles of a triangle are equal to 180°, and our triangle is isosceles, so half the remaining amount (after subtracting angle A) is equal to the two acute angles (ABE and BEA).

1/2(180°-108° = 72°)

= 36°