Suppose that the position of point A after folding is A', connecting AA'.

DA = DA', AA' is perpendicular to DE,

Angle 1= angle da' a+ angle da' = 2 * angle da 'a.

Angle 2=2* Angle EA 'a

Angle 1+ angle 2=2* (angle da' a+ angle ea' a) = 2 * angle a.

Reply zhangrq_qd: That makes sense. It is proved here that AA' is perpendicular to DE:

Because it is folded, it must have:

Triangle EDA is congruent with triangle EDA',

So, angle EDA= angle EDA',

Connect AA' intersection point DE to point f,

Obviously, there are ad = a'd and ***DF edges,

Therefore, the triangle da'f is congruent with the triangle da'f,

So af = a' f, that is, point f is the midpoint of AA',

And AD=A`D,

So DF is perpendicular to AA', that is, DE is perpendicular to AA'.

Because you can see it at a glance, there is no proof in the earliest answer. pity