Bisector of Foci Angle (Posted on 2015-08-05)

	Submitted by Bractals
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Let F and F' be the foci of Γ. We will add the following line segments to our drawing : FA, AF', FB, and BF'. ------------------------------------------------------------------------- Construct A1 and B1 as the reflections of F about lines AC and BC respectively, Join F to A1 and B1 and B1 to A1 and B. Let A2 and B2 be the midpoints of FA1 and FB1 respectively. Join A to A1 and A2 and B to B2. Clearly, ΔA1A2A ≅ ΔFA2A and ΔB1B2B ≅ ΔFB2B. Therefore, AA1 = AF and ∠A1AA2 = ∠FAA2,                    BB1 = BF and ∠B1BB2 = ∠FBB2. The equality of the angles and the reflective property of ellipses imply that F', A, and A1 are collinear; likewise for F', B, and B1.      F'A1 = F'A + AA1               = F'A + AF               = F'B + BF   ( from def. of ellipse )               = F'B + BB1               = F'B1 Thus, A1F'B1 is an isosceles triangle. Point C is the intersection of the perpendicular bisectors of FA1 and FB1. Therefore, It is the circumcenter of ΔFA1B1. The perpendicular bisector of A1B1 passes through C and F'. It also bisects ∠A1F'B1 = ∠AF'B. ------------------------------------------------------------------------- We have proved the proposition for focus F'. To prove it for focus F, repeat everything between the lines swapping F for F' and F' for F. QED