Prove the two corollaries given by products of proposition I
Solution
1. For n sided polygon we can divide it in n traingles by taking any point P inside polygon and drawing straight lines to eah vertex.
Now angle of these traingles will be 2*90 (twice the right angle). Combining all these n traingles we have 2n*90 (2n times right angle) which include interior angles + all angles at that point P.
sum of all angle at P = 4*90 (4 times right angle)
Hence sum of interior angles = (2n-4) right angles
2. Divide a polygon of n sides into triangles such that these traingles are made by non-intersecting diagonals between the vertices of the polygon. We will end up in getting n-2 such traingles for n sided polygon.
Hence internal angle = (n-2)*180 = 180n - 360
Also, n sided polygon has n vertices which will make angle = 180*n
Hence sum of exterior angle = 180n - (180n - 360) = 360 (4 right angles)
