Question

Prove that each diagonals of a regular sepatagon is parallel to one and only one side.



Please prove it please its humble request. ​

Answer 1

Construction: in the regular hexagon ABCDEF, join FC.

Construction: in the regular hexagon ABCDEF, join FC.In the quadrilateral ABCF,AF=BC and ∠FAB=∠CBA=120

Construction: in the regular hexagon ABCDEF, join FC.In the quadrilateral ABCF,AF=BC and ∠FAB=∠CBA=120 0

Construction: in the regular hexagon ABCDEF, join FC.In the quadrilateral ABCF,AF=BC and ∠FAB=∠CBA=120 0

Construction: in the regular hexagon ABCDEF, join FC.In the quadrilateral ABCF,AF=BC and ∠FAB=∠CBA=120 0 ABCF is an isosceles trapezium

Construction: in the regular hexagon ABCDEF, join FC.In the quadrilateral ABCF,AF=BC and ∠FAB=∠CBA=120 0 ABCF is an isosceles trapeziumFC∣∣AB…….(1)

Construction: in the regular hexagon ABCDEF, join FC.In the quadrilateral ABCF,AF=BC and ∠FAB=∠CBA=120 0 ABCF is an isosceles trapeziumFC∣∣AB…….(1)Similarly ED∣∣FC…….(2)

Construction: in the regular hexagon ABCDEF, join FC.In the quadrilateral ABCF,AF=BC and ∠FAB=∠CBA=120 0 ABCF is an isosceles trapeziumFC∣∣AB…….(1)Similarly ED∣∣FC…….(2)From (1) and (2), we get AB∣∣ED

Construction: in the regular hexagon ABCDEF, join FC.In the quadrilateral ABCF,AF=BC and ∠FAB=∠CBA=120 0 ABCF is an isosceles trapeziumFC∣∣AB…….(1)Similarly ED∣∣FC…….(2)From (1) and (2), we get AB∣∣EDSimilarly, BC∣∣FE and CD∣∣FA.

Construction: in the regular hexagon ABCDEF, join FC.In the quadrilateral ABCF,AF=BC and ∠FAB=∠CBA=120 0 ABCF is an isosceles trapeziumFC∣∣AB…….(1)Similarly ED∣∣FC…….(2)From (1) and (2), we get AB∣∣EDSimilarly, BC∣∣FE and CD∣∣FA.Hence, the opposite sides of a regular hexagon are parallel.