Hexagon ABCDEF is a regular hexagon. Show that segments AE and ED are perpendicular.
1 answer:
For a regular hexagon, the interior angles are 120°, since they add up to 720 (180(6 - 2))
We also know, for a regular hexagon, all sides are equal.
Refer to my diagram.
Triangle AFE is an isosceles triangle, because two sides of the triangle are equal in a regular hexagon.
Thus, 120 + 2x = 180 (angle sum of triangle)
2x = 60 and x = 30
Now, ∠FEA = 30°, but we know that ∠FED is 120° (all interior angles of a regular hexagon is 120°)
We know that ∠FED = ∠FEA + ∠AED
120° = 30° + ∠AED
∴ ∠AED = 90° (120 - 30)
If ∠AED is 90°, then, by definition, AE ⊥ ED
We also know, for a regular hexagon, all sides are equal.
Refer to my diagram.
Triangle AFE is an isosceles triangle, because two sides of the triangle are equal in a regular hexagon.
Thus, 120 + 2x = 180 (angle sum of triangle)
2x = 60 and x = 30
Now, ∠FEA = 30°, but we know that ∠FED is 120° (all interior angles of a regular hexagon is 120°)
We know that ∠FED = ∠FEA + ∠AED
120° = 30° + ∠AED
∴ ∠AED = 90° (120 - 30)
If ∠AED is 90°, then, by definition, AE ⊥ ED
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