Draw a square and a circle through all four vertices. Draw diameters parallel to the sides of the square and draw a polygon joining the end points of these diameters and the vertices of the square.


Prove that this polygon is a regular octagon.


D is the centre of the circle. EFGH is the inscribed square.


Let, side of the square = a


∴ EF = FG = GH = HE = PM = NO = a


∴ Diagonal of the square = a√2


∴ EG = a√2


Now we have, EG = IJ = KL [diameter of the circle]


∴ EG = IJ = KL = a√2


IJ = a√2 and PM = a


[∵IM = PJ]



In ΔIME,


EM = a/2 [∵ EM = MF = EF/2]



∠IME = 90°


∴ tan ∠MIE = EM/IM




∴ tan ∠MIE = √2 + 1


∴ tan ∠MIE = tan 67.5°


∴ ∠MIE = 67.5°……… (1)


Similarly in ΔIMF,


∠MIF = 67.5°


∴ Inner angle of the polygon,


⇒ ∠EIF = ∠MIE + ∠MIF = 67.5°+67.5° = 135°


Similarly all the inner angles of the polygon = 135°


∴ All the outer angles of the Polygon = 180° - 135° = 45°


∴ It is a regular polygon.


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