Given:
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1.
Segment EC = segment CD which is given.
2.
Angle CEA = angle CDB by angle complements.
3.
Segment EA = segment DB given from ABDE
being a unit square.
4.
Therefore, triangle AEC is congruent to triangle BCD
by SAS.
5.
Segment AC = segment CB by CPCTC.
1. Since the altitude of a unit equilateral triangle will be a perpendicular bisector of the base, the distance from any base angle to the altitude is 0.5.
2. Since the triangle formed by the constructed altitude and either base angle 60o each is a right triangle, we can use right-triangle trigonometry to solve for the altitude h.
3. Tangent 60o = h/0.5
4. Hence h = 0.5(tan(60o)) = 0.8660254038
1. Construct perpendicular line segment GH through point C that is the perpendicular bisector to DE.2. This line will be the perpendicular bisector of ED since triangle CDE is isosceles. Segment GH will also be perpendicular to AB since a line that is perpendicular to one of two parallel lines all in the same plane will be perpendicular to the 2d parallel line.
3. Find the altitude CG: tan 15o = CG/0.5 or CG = 0.5(tan15o) = 0.1339745962
4. Find segment CH: CH = 1 - 1.339745962 = 0.8660254038
5. Since segment CH = 0.8660254038, and by our lemma we know that the altitude of any unit equilateral triangle is 0.8660254038, then we can conclude that ABC is an equilateral triangle.
QED