Question: Straight line with fixed direction

We give a line (D) and a point A located at a distance AH=h from D. A constant angle of magnitude alpha pivots to its apex A and we call B and C the points where its sides cut the line D. Let O be the center of the circle circumscribed to the triangle ABC.
Demonstrate that the B and C tangents to the O circle keep a fixed direction. 
Here is my code which don't work for slopes are not equal.

restart; with(plots): with(geometry):unprotect(D):
_EnvHorizontalName := 'x':_EnvVerticalName := 'y':
line(D, y = (1/2)*x-1); point(A, 5, 5); PerpendicularLine(lp, A, D); h := distance(A, D); intersection(H, D, lp);
alpha := (1/16)*Pi;
rotation(lp1, lp, (1/6)*Pi, 'clockwise', A); rotation(lp2, lp1, (1/6)*Pi-alpha, 'clockwise', A); FindAngle(lp1, lp2); evalf(%);
intersection(B, D, lp1); intersection(C, D, lp2);
triangle(T, [A, B, C]);
circumcircle(Elc, T, 'centername' = OO);
TangentLine(tgB, B, Elc); TangentLine(tgC, C, Elc);
evalf(slope(tgB)); evalf(slope(tgC));
dr := draw([D(color = blue), lp(color = red), Elc(color = green), A, B, C, T(color = black), H, tgB, tgC], printtext = true);

display([dr], axes = none, scaling = constrained);
Fig := proc (k) local dr, Elc, B, C, lp1, lp2; global D, A, lp, H, alpha; geometry:-rotation(lp1, lp, (1/6)*Pi+k, 'clockwise', A); geometry:-rotation(lp2, lp1, (1/6)*Pi-alpha+k, 'clockwise', A); geometry:-intersection(B, D, lp1); geometry:-intersection(C, D, lp2); geometry:-triangle(T, [A, B, C]); geometry:-circumcircle(Elc, T, 'centername' = OO); geometry:-TangentLine(tgB, B, Elc); geometry:-TangentLine(tgC, C, Elc); dr := geometry:-draw([D(color = blue), lp(color = red), Elc(color = green), A, B, C, T(color = black), H, tgB, tgC], printtext = true); plots:-display([dr], axes = none, scaling = constrained) end proc;
iframes := 10;

display([seq(Fig((1/12)*Pi+i/(10*iframes)), i = 1 .. iframes)], insequence, scaling = constrained);
How to improve this code ? Thank you.