.geometry "version 0.31";
.l0; 
v1 = .free(-0.260864, 0.355724, "A");
v2 = .free(-0.475733, 0.595865, .in, "B");
v3 = .free(0.394261, 0.254936, "C");
v4 = .free(0.601076, 0.638595, .in, "D");
c1 = .c.vv(v1, v2);
c2 = .c.vv(v3, v4);
v5 = .v.cc(c1, c2, 1, "E");
l1 = .l.vc(v5, c2, 2, .longline);
l2 = .l.vc(v5, c1, 2, .longline);
v6 = .v.lc(l1, c1, 1, "F");
v7 = .v.lc(l2, c2, 2, "G");
c3 = .c.vvv(v5, v7, v6);
l3 = .l.vc(v5, c3, 2, .longline);
v8 = .v.ccenter(c3, [.in, .blink, .white], "H");
v9 = .v.lc(l3, c2, 2, "I");
v10 = .v.lc(l3, c1, 1, "J");
l4 = .l.vv(v1, v8, [.in, .blink, 2 .green, 2 .blink1, .green]);
l5 = .l.vv(v8, v3, [.in, .blink, .red, .blink2, .red, .blink, .red]);
l6 = .l.vv(v3, v5, [.in, .blink, 2 .blue, 2 .blink1, .blue]);
l7 = .l.vv(v5, v1, [.in, .blink, .blue, .blink2, .blue, .blink, .blue]);
l8 = .l.vv(v8, v5, [9 .in, .blink, .blue]);
l9 = .l.vv(v3, v9, [5 .in, .blink1, .green]);
l10 = .l.vv(v1, v10, [5 .in, .blink, .red]);
l11 = .l.vv(v10, v8, [8 .in, .blink, .yellow]);
l12 = .l.vv(v8, v9, [8 .in, .blink, .yellow]);
.text("Two circles centered at A and C meet at
E.  Let EF and EG be the tangent lines at E that
meet the circles at F and G.  Construct a
circle passing through E, F, and G, and let
IJ be the tangent to the new circle at E.
Show that EI = EJ.", .l0);
.text("Let H be the center of the new circle.
We will begin by showing that
EAHC is a parallelogram.", .l1);
.text("Since EG is tangent to the circle
centered at A and EA is a radius, EA\perp EG.", .l2);
ang1 = .a.vvv(v7, v5, v1, .blink, .l2, .right);
v11 = .v.ll(l2, l5, .in, .l3, "A");
ang2 = .a.vvv(v7, v11, v8, .blink, .l3, .right);
.text("Since HC connects the centers of the
circles whose common chord is EG, HC\perpEG.
Since both AE and HC are perpendicular to EG,
AE and HC are parallel.", .l3);
.text("Exactly the same argument can be used
to show that AH is parallel to EC, so
EAHC is a parallelogram.", .l4);
.text("Thus HC = AE and since AE and AI
are radii of the same circle, HC = AE = AI.
Similarly, AH = EC = CI.", .l5);
.text("We want to show that \angleHCI = \angleHAJ.
This will be done by showing that
\angleHCI (exterior) + \angleHAJ = 360\degrees.", .l6);
v12 = .v.lc(l7, c1, 2, .blink, .l7, "X");
l13 = .l.vv(v12, v1, .blink, .l7);
.text("\angleXAJ = 2\angleAEJ, \angleICY = 2\angleIEY,
and \angleHCY = \angleAEC = \angleYAH, so
\angleHCY + \angleICY + \angleHAJ =
\angleHCY + \angleXAH + \angle ICY + \angleXAJ
= 2\angleIEY + 2\angleAEC + 2\angleAEJ = 360\degrees.", .l7);
v13 = .v.lc(l6, c2, 1, .blink, .l7, "Y");
l14 = .l.vv(v13, v3, .blink, .l7);
.text("Thus \angleHAJ = \angleHCI, so by SAS,
\triangleHAJ \congruent \triangleICH, so HI = HJ", .l8);
.text("HE \perp IJ, HE = HE, and HI = HJ, so by
hypotenuse-leg, \triangleHEI \congruent \triangleHEJ.
Thus EI = EJ, which is what we wanted to show.", .l9);
.text("Press 'Next' to continue ...", .red, .tol8);
ang3 = .a.vvv(v8, v1, v10, .blink2, .l6);
ang4 = .a.vvv(v9, v3, v8, .blink2, .l6, .ring2);