Intersection of Bisectors (Posted on 2012-12-01)

	Submitted by Bractals
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Why the restriction that chord AC does not contain point O? If it did, then ∠AB1C = ∠AB2C = 90° and therefore H1 = B1 and H2 = B2. What then is the meaning of the bisector of ∠B1B1O and ∠B2B2O? WOLOG assume that ∠AB1C > 90°. From Coincide or Perpendicular we know that the bisector of ∠H1B1O is perpendicular to the bisector of ∠AB1C. There are two cases: Case I: The bisector of ∠H1B1O is tangent to Γ. In this case line B1B2 is the perpendicular bisector of chord AC. The bisector of ∠H2B2O (which coincides with the bisector of ∠AB2C) is ray B2B1. Therefore, the bisectors intersect at point B1. Case II: The bisector of ∠H1B1O is not tangent to Γ. Let the bisector of ∠H1B1O intersect Γ again at point P1. Let the bisector of ∠AB1C intersect Γ again at point P2. Since these bisectors are perpendicular, line segment P1P2 contains point O. We will show that ray B2P1 bisects ∠H2B2O. First construct line segments OB1 and OB2.      ∠B2B1P2 + ∠P2B1P1 + ∠P1B1H1 = ∠B2B1H1 = 2*∠P2B1P1   ⇒ ∠B2B1P2 + ∠P1B1H1 = ∠P2B1P1   ⇒ ∠B2B1P2 + ∠OB1P1 = ∠P2B1O + ∠OB1P1   ⇒ ∠B2B1P2 = ∠P2B1O                                   (*)      ∠H2B2P1 = ∠B1B2P1              = ∠B1P2P1              = ∠B1P2O              = ∠P2B1O              = ∠B2B1P2                    from (*)              = ∠B2P1P2              = ∠B2P1O              = ∠P1B2O   ⇒ B2P1 bisects ∠H2B2O From the above we have      ∠B1P2P1 = ∠B2B1P2   ⇒ P1P2 || B1B2   ⇒ P1P2 ⊥ AC   ⇒ P1P2 is the perpendicular bisector of AC with P1 in arc (AB1C). Therefore, the intersection of the bisector of ∠H1B1O and the bisector of ∠H2B2O is the point P1 which is independent of the point D on chord AC. QED

	Subject	Author	Date
	Proof.	Jer	2012-12-03 13:56:48
	From sketchpad. No proof.	Jer	2012-12-03 13:10:40