EMAT 8890
Doug Westmoreland
Hint for Brahmagupta's Theorem:
In the GSP sketch, let the acute angles in right triangle CGP be x and y. Since the diagonals are perpendicular, angle DPG = x, and in right triangle DGP the angle at D is y. The pairs of vertically opposite angles at P then give further angles x and y. But the chord AD subtends equal angles x at C and B on the circumference, and chord BC similarly gives equal angles y at A and D. Thus the triangles PFA and PFB are both isosceles, and the AF and FB are equal.