# In the figure below,ABCDEF is a regular hexagon.prove that ABDE is a rectangle

In Triangles, AFE and BCD,

AF = BC (sides of regular hexagon are equal)

angle AFE = angle BCD (angles of regular hexagon are equal)

EF = DC (sides of regular hexagon are equal)

so, triangle AFE congruent triangle BCD

so, AE = BD (CPCT)

again, AB = DE (sides of regular hexagon are equal)

so, ABDE is a parallelogram

Now, in triangle AFE,

angle AFE + FAE + FEA = 180 (angle sum property of triangle)

or 120 + 2 angle FAE = 180 [all angles of regular hexagon = 120 and FAE = FEA as AF = FE]

or 2 angle FAE = 180 - 120 = 60

or angle FAE = 60 / 2 = 30

so, angle BAE = angle BAF - FAE

= 120 - 30 = 90

so ABDE is a rectangle (proved)