.geometry "version 0.51";
.l0; 
.text("Construct a triangle given the measure of
one angle (\alpha), the length of the internal bisector
of that angle (L\sub{A}, and the radius (r) of the incircle.", .l0);
v1 = .free(-0.616766, 0.859281);
v2 = .free(-0.844311, 0.694611);
v3 = .free(-0.5, 0.691617);
v4 = .free(-0.841317, 0.598802);
v5 = .free(-0.266467, 0.598802);
v6 = .free(-0.829341, 0.491018);
v7 = .free(-0.694611, 0.48503);
l1 = .l.vv(v1, v2);
l2 = .l.vv(v2, v3);
l3 = .l.vv(v4, v5, "L\sub{A}");
l4 = .l.vv(v6, v7, "r");
ang1 = .a.vvv(v3, v2, v1, "\alpha");
v8 = .free(-0.233533, -0.00598802, "A");
v9 = .free(0.742515, -0.00898204, .in);
l5 = .l.vv(v8, v9, [.in, .white], .ray12);
v10 = .v.avv(ang1, v9, v8, .in);
l6 = .l.vv(v8, v10, [.in, .white], .ray12);
v11 = .v.vvvbisect(v9, v8, v10, .in);
l7 = .l.vv(v8, v11, [.in, 4 .white, .in], .ray12);
c1 = .c.ctrvv(v8, v4, v5, [2 .in, .white, 4 .in, .white]);
l8 = .l.vlperp(v9, l5, .in, .ray12);
c2 = .c.ctrvv(v9, v6, v7, .in);
v12 = .v.lc(l8, c2, 1, .in);
l10 = .l.vlpar(v12, l5, [3 .in, .white], .longline);
v13 = .v.ll(l7, l10, [.white, 3 .in, .blink, .white]);
c3 = .c.ctrvv(v13, v6, v7, [.white, 4 .in, .blink, .white]);
v14 = .v.lc(l7, c1, 2, [.white, .in, .white]);
l11 = .l.vc(v14, c3, 1, [6 .in, .blink, .white], .longline);
v15 = .v.ll(l6, l11, [.white, 5 .in, .blink, .white], "B");
v16 = .v.ll(l5, l11, [.white, 5 .in, .blink, .white], "C");
l12 = .l.vlperp(v13, l5, .in);
v17 = .v.ll(l5, l12, [.white, 3 .in, .white]);
l13 = .l.vv(v13, v17, [.white, 3 .in, .white], "r");
l9 = .l.vv(v8, v16, [.white, .in]);
l14 = .l.vv(v16, v15, [.white, .in]);
l15 = .l.vv(v15, v8, [.white, .in]);
l16 = .l.vv(v14, v8, [.white, .in], "L\sub{A}");
.text("Copy angle \alpha and construct its angle
bisector.", .l1);
.text("Draw a circle of radius L\sub{A} about
the vertex.  Its intersection with the angle bisector
is the point where the angle bisector meets the
opposite side.", .l2);
.text("Construct a line parallel to the base a
distance of r from it.  The center of the incircle
must lie on this line.", .l3);
.text("The center of the incircle must also lie on
the angle bisector, so find the intersection of the
angle bisector and the parallel line for the center
of the incircle.", .l4);
.text("Draw the incircle centered at the point
we just constructed.", .l5);
.text("The tangent to the incircle from the endpoint
of L\sub{A} is the third side of the triangle, so
we are done.", .l6);
.text("Press 'Next' to continue ...", .red, .tol5);