VERY DIFFICULT GEOMETRY Q2

Draw a regular dodecagon with length 4 (from center O) onto a rectangular coordinate plane.
Such that vertexes P1, P2, P3 in quadrant I, P4, P5, P6 in quadrant II, ...
So P1 is (4cos75, 4sin75) = (4cos(45 + 30), 4sin(45 + 30)) = (√6 - √2, √6 + √2)
P2 is (4cos45, 4sin45) = (2√2, 2√2)
P3 is (4cos15, 4sin15) = (4cos(45 - 30), 4sin(45 - 30)) = (√6 + √2, √6 - √2)
. . .
Let the coordinates of the said square (in question) ABCD be
A(2, 2), B(2, -2), C(-2, -2), D(-2, 2).
So the points L, M, N, K are (0, 2(√3 - 1)), (2(√3 - 1), 0), (0, 2(1 - √3)), (2(1 - √3, 0) respectively.
Mid-pt of AK, LM, LB are Q1, Q2, Q3 respectively. Therefore,
Q1 = (1/2)AK = (2 - √3, 1);
Q2 = (1/2)LM = (√3 - 1, √3 - 1);
Q3 = (1/2)AN = (1, 2 - √3).

slope of OQ1 = 1/(2 - √3) = 2 + √3, and,
slope of OP1 = (√6 + √2)/(√6 - √2) = 2 + √3 = slope of OQ1;
slope of OQ2 = 1, and slop of OP2 = 1 = slop of OQ2;
slope of OQ3 = 2 - √3, and,
slope of OP3 = (√6 - √2)/(√6 + √2) = 2 - √3 = slope of OQ3.

OQ1 : OP1
= √[(2 - √3)^2 + 1] : √[(√6 - √2)^2 + (√6 + √2)^2]
= √(8 - 4√3) : √[2(6 + 2)]
= √(2 - √3) : 2
OQ2 : OP2
= √[(√3 - 1)^2 + (√3 - 1)^2)] : √[(2√2)^2 + (2√2)^2]
= √(8 - 4√3) : √16
= √(2 - √3) : 2
OQ3 : OP3
= √[1 + (2 - √3)^2] : √[(√6 + √2)^2 + (√6 - √2)^2]
= √(8 - 4√3) : √[2(6 + 2)]
= √(2 - √3) : 2

As OQ1 : OP1 = OQ2 : OP2 = OQ3 : OP3 = √(2 - √3) : 2
and slope of OQ1 = slope of OP1, slope of OQ2 = slope of OP2, slope of OQ2 = slope of OP2.
Similarly to vertexes Q4, Q5, Q6, ..., Therefore,
Q1, Q2, Q3, ... is a regular dodecagon.