Consider this regular pentagon #ABCDE#.
Let us join vertices #AC# and #EC# as shown to form three triangles as shown. I have used letters #a, b, c, d, e, f, g, h, i# to represent internal angles of triangles for sake of simplicity.
Since the sum of interior angles of a triangle is #180^o#,
In #triangleABC, b+c+d = 180^o#
In #triangleACE, a+e+i = 180^o#
In #triangleECD, h+f+g = 180^o#
Sum of interior angles of the pentagon is
#a+b+c+d+e+f+g+h+i#
#=(b+c+d)+(a+e+i)+(h+f+g)#
#=180^o + 180^o +1 80^o# [using the above three results]
#=540^o#
#i.e. angleA+angleB+angleC+angleD+angleE=540^o#
Since it is a regular octagon, #angleA=angleB=angleC=angleD=angleE#
#implies angleA+angleA+angleA+angleA+angleA = 540^o#
#implies 5*angleA = 540^o#
#implies angleA=540/5=108^o = angleB=angleC=angleD=angleE#
Hence internal angle of a regular pentagon is #108^o#.
