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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.
