Prove that in a regular pentagon the diagonals outgoing from one vertex divide the angle

Prove that in a regular pentagon the diagonals outgoing from one vertex divide the angle at the given vertex into three equal parts

Given:
ABCDE – regular pentagon
AB = BC = CD = DE = AE
<ABC = <BCD = <CDE = <DEA = <EAB = 108 degrees.
AC, AD-diagonals
Prove:
<BAC = <CAD = <DAE
Solution:
Consider triangles ABC and DEA, in them, AB = BC, AE = DE, since these are the sides of the pentagon, it turns out that these triangles are isosceles. Also, <ABC = <DEA, which follows from the problem statement. Triangle ABC = triangle DEA on two sides and the angle between them.

Then <BAC = <CAD and AC = AD. In triangle CAD AC = AD, it is isosceles.
<BAC = (180-108) / 2 = 36
<CAD = 108-36-36 = 36, <CAD = <DAE = <BAC as required.