Inscribed Triangle

What is the length of the side of an equailateral triangle inscribed in a circle of radius 1?

StatementReason
Triangle ABD is inscribed in the circle with center at C. Construction
AC, BC, and DC are all length 1 They are all radii of the circle
If we continue DC until it intersects AB at E, the two lines meet at right angles Since the triangle is equilateral, DC bisects angle ADB and so the two angles at E are equivalent and add up to 180 degrees (*)
Angle CAE is 30 degress AC bisects angle DAB
AE is length sqrt(3)/2 It's part of a 30-60-90 right triangle
AD is length sqrt(3) AD is twice the length of AE

(*) - I left out a few steps from a completely formal proof, but you get the idea here