We show that the interior angles are 108°
and that they are trisected at each vertex
into three 36° angles.

The sum of the interior angles is 

(n-2)*180° = (5-2)*180° = 3*180° = 540°

Since they are all equal, each interior
angle is 108°

Angle ABC = 108°
Triangle ABC is isosceles because AB = BC.
The base angles are (180°-108°)/2 = 36° each.
So Angle BAC = BCA = 36°

Angle AED = 108°
Triangle AED is isosceles because AE = ED.
The base angles are (180°-108°)/2 = 36° each.
So Angle EAD = DEA = 36°

Angle BAC = 108° = BAC+EAD+CAD
      BAC = 108° = 36°+36°+CAD
            108° = 72°+CAD
             36° = CAD

So each of the angles at A are 36°.  That is easily
proved at all 5 vertices.

Therefore it's easy to show that the base angles
of both triangles ABD and AEC are 72° (making them
isosceles) and the corresponding sides between them 
are AB and BE respectively, so they are congruent 
by angle-side-angle.

Edwin