Coincide or Perpendicular (Posted on 2012-10-20)

	Submitted by Bractals
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The following assumptions were made when constructing the drawings for this problem: 1) side AC will be horizontal with A to the left and C to the right, 2) vertex B will lie above line AC, and 3) ∠A < ∠C. Construct the circumcircle Γ of ΔABC and ray BO intersecting Γ again at point D. Construct line BH intersecting line AC at point E. Construct the bisector of ∠ABC intersecting side AC at point F. Construct line segment HO and the bisector of ∠HBO intersecting it at point G. Construct line segments AD and CD. Playing around with Geometer's Sketchpad it was found that the bisectors coincide if ∠B < 90° and they are perpendicular if ∠B > 90°. Four cases will be analyzed for the proof: Case I: ∠B and ∠C < 90° Here we have B•H•E (see definition below) and ABCD a cyclic quadrilateral with diagonals AC and BD.     ∠ADB = ∠ACB = ∠ECB and     ∠BAD = ∠BEC   ⇒ ΔBAD ~ ΔBEC   ⇒ ∠ABD = ∠EBC      BF bisects ∠ABC   ⇒ ∠ABF = ∠FBC   ⇒ ∠ABD + ∠DBF = ∠FBE + ∠EBC   ⇒ ∠DBF = ∠FBE   ⇒ ∠OBF = ∠FBH   ⇒ BF bisects ∠HBO Case II: ∠C = 90° Here O is the midpoint of side AB and H = C. BF bisects ∠ABC and therefore it bisects ∠HBO. Case III: ∠C > 90° Here we have B•E•H and ACBD a cyclic quadrilateral with diagonals AB and CD.     ∠BCD = ∠BEC     ∠BCA = ∠BEC + ∠CBE   ⇒ ∠BCD + ∠DCA = ∠BEC + ∠CBE   ⇒ ∠DCA = ∠CBE   ⇒ ∠DBA = ∠CBE      BF bisects ∠ABC   ⇒ ∠ABF = ∠FBC   ⇒ ∠DBF - ∠DBA = ∠FBE - ∠CBE   ⇒ ∠DBF = ∠FBE   ⇒ ∠OBF = ∠FBH   ⇒ BF bisects ∠HBO Case IV: ∠B > 90° Here we have H•B•E and ABCD a cyclic quadrilateral with diagonals AC and BD.     ∠BCE + ∠ACD = 90° = ∠BCE + ∠EBC   ⇒ ∠ACD = ∠EBC      BF bisects ∠ABC   ⇒ ∠ABD = ∠FBC   ⇒ ∠ABD + ∠DBF = ∠FBE + ∠EBC   ⇒ ∠ACD + ∠DBF = ∠FBE + ∠ACD   ⇒ ∠DBF = ∠FBE     ∠HBG + ∠GBO + ∠OBF + ∠FBE = ∠HBE = 180°   ⇒ ∠GBO + ∠GBO + ∠DBF + ∠DBF = 180°   ⇒ ∠GBO + ∠DBF = 90°   ⇒ ∠GBF = 90°   ⇒ BF ⊥ GF QED DEFINITION   P•Q•R denotes collinear points P, Q, and R   with Q between P and R.